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A. यह लंबी दूरी के बमवर्षकों के लिए सहायक और आपात अड्डा बना/It became a support and emergency base for long-range bombers
Step 1
Concept
Iwo Jima was a strategic base for air war near Japan. For exams understand islands as military bases.
Step 2
Why this answer is correct
The correct answer is A. यह लंबी दूरी के बमवर्षकों के लिए सहायक और आपात अड्डा बना / It became a support and emergency base for long-range bombers. Iwo Jima was a strategic base for air war near Japan. For exams understand islands as military bases.
Step 3
Exam Tip
इवो जीमा जापान के निकट वायु युद्ध में रणनीतिक अड्डा था। परीक्षा में द्वीपों का उपयोग सैन्य अड्डों के रूप में समझें।
Changing all signs in the second bracket gives \(2y^3-y+5-5y^3-4y+8\). In exams, change the sign of every term during subtraction.
Step 2
Why this answer is correct
The correct answer is A. \(,-3y^3-5y+13,\). Changing all signs in the second bracket gives \(2y^3-y+5-5y^3-4y+8\). In exams, change the sign of every term during subtraction.
Step 3
Exam Tip
दूसरे bracket के सभी signs बदलने पर \(2y^3-y+5-5y^3-4y+8\) मिलता है। परीक्षा में subtraction में हर पद का sign बदलें।
Changing the signs of the second bracket gives \(5x^3-2x+7-2x^3-3x+5\), so the answer is \(3x^3-5x+12\). In exams, change the sign of every term in the bracket during subtraction.
Step 2
Why this answer is correct
The correct answer is A. \(,3x^3-5x+12,\). Changing the signs of the second bracket gives \(5x^3-2x+7-2x^3-3x+5\), so the answer is \(3x^3-5x+12\). In exams, change the sign of every term in the bracket during subtraction.
Step 3
Exam Tip
दूसरे bracket के signs बदलकर \(5x^3-2x+7-2x^3-3x+5\) मिलता है, इसलिए उत्तर \(3x^3-5x+12\) है। परीक्षा में subtraction में पूरे bracket का sign बदलें।
A. क्योंकि वे व्यवहार में विकसित शांति प्रबंधन का साधन बने/Because they developed in practice as a tool of peace management
Step 1
Concept
Peacekeeping developed from the practical role of the UN. Exam tip: connect it with the peace spirit of the Charter.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि वे व्यवहार में विकसित शांति प्रबंधन का साधन बने / Because they developed in practice as a tool of peace management. Peacekeeping developed from the practical role of the UN. Exam tip: connect it with the peace spirit of the Charter.
Step 3
Exam Tip
शांति स्थापना संयुक्त राष्ट्र की व्यावहारिक भूमिका से विकसित हुई। परीक्षा में इसे चार्टर की शांति भावना से जोड़ें।
A. संयुक्त राष्ट्र शांति सैनिकों से/United Nations peacekeepers
Step 1
Concept
Blue helmets identify UN peacekeepers. For exams connect them with peacekeeping.
Step 2
Why this answer is correct
The correct answer is A. संयुक्त राष्ट्र शांति सैनिकों से / United Nations peacekeepers. Blue helmets identify UN peacekeepers. For exams connect them with peacekeeping.
Step 3
Exam Tip
नीले हेलमेट संयुक्त राष्ट्र शांति सैनिकों की पहचान हैं। परीक्षा में इन्हें शांति स्थापना से जोड़ें।
A. शांति स्थापना संघर्ष रोकने और स्थिरता में सहायता करती है विजय नहीं चाहती/Peacekeeping helps stop conflict and support stability not conquest
Step 1
Concept
UN peacekeepers support peace processes and civilian protection. Exam tip: remember them by blue helmets.
Step 2
Why this answer is correct
The correct answer is A. शांति स्थापना संघर्ष रोकने और स्थिरता में सहायता करती है विजय नहीं चाहती / Peacekeeping helps stop conflict and support stability not conquest. UN peacekeepers support peace processes and civilian protection. Exam tip: remember them by blue helmets.
Step 3
Exam Tip
संयुक्त राष्ट्र शांति सैनिक शांति प्रक्रिया और नागरिक सुरक्षा में सहायता करते हैं। परीक्षा में उन्हें नीली टोपी से याद करें।
A. यदि जमीनी सेना जल्दी न पहुंचे तो हवाई सैनिक अलग पड़ सकते हैं/Airborne troops can become isolated if ground forces do not arrive quickly
Step 1
Concept
In Market Garden dependence on bridges and narrow routes became risky. For exams study both bold plan and ground reality.
Step 2
Why this answer is correct
The correct answer is A. यदि जमीनी सेना जल्दी न पहुंचे तो हवाई सैनिक अलग पड़ सकते हैं / Airborne troops can become isolated if ground forces do not arrive quickly. In Market Garden dependence on bridges and narrow routes became risky. For exams study both bold plan and ground reality.
Step 3
Exam Tip
मार्केट गार्डन में पुलों और संकरे मार्गों पर निर्भरता जोखिम बनी। परीक्षा में साहसिक योजना और जमीन की वास्तविकता दोनों पढ़ें।
A. यह जापान के निकट आपात और सहायक वायु अड्डे के रूप में उपयोगी था/It was useful as an emergency and support air base near Japan
Step 1
Concept
Iwo Jima had strategic value for air operations near Japan. For exams see islands not only as land but as bases.
Step 2
Why this answer is correct
The correct answer is A. यह जापान के निकट आपात और सहायक वायु अड्डे के रूप में उपयोगी था / It was useful as an emergency and support air base near Japan. Iwo Jima had strategic value for air operations near Japan. For exams see islands not only as land but as bases.
Step 3
Exam Tip
इवो जीमा जापान के निकट वायु अभियानों में सामरिक महत्व रखता था। परीक्षा में द्वीपों को केवल भूमि नहीं बल्कि अड्डों के रूप में देखें।
Here \(\sqrt{363}=11\sqrt{3}\), \(2\sqrt{147}=14\sqrt{3}\), and \(3\sqrt{75}=15\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value should be (12).
Step 2
Why this answer is correct
The correct answer is C. (15). Here \(\sqrt{363}=11\sqrt{3}\), \(2\sqrt{147}=14\sqrt{3}\), and \(3\sqrt{75}=15\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value should be (12).
Step 3
Exam Tip
\(\sqrt{363}=11\sqrt{3}\), \(2\sqrt{147}=14\sqrt{3}\), और \(3\sqrt{75}=15\sqrt{3}\)। अंश \(12\sqrt{3}\) है, इसलिए मान (12) होना चाहिए।
Here (\left\(\frac{5}{8}\right\)^{-2}=\frac{64}{25}) and (\left\(\frac{8}{5}\right\)^{-2}=\frac{25}{64}). The sum is \(\frac{4096+625}{1600}=\frac{4721}{1600}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4721}{1600}\). Here (\left\(\frac{5}{8}\right\)^{-2}=\frac{64}{25}) and (\left\(\frac{8}{5}\right\)^{-2}=\frac{25}{64}). The sum is \(\frac{4096+625}{1600}=\frac{4721}{1600}\).
Step 3
Exam Tip
(\left\(\frac{5}{8}\right\)^{-2}=\frac{64}{25}) और (\left\(\frac{8}{5}\right\)^{-2}=\frac{25}{64})। योग \(\frac{4096+625}{1600}=\frac{4721}{1600}\) है।
From \(\sqrt{x}=5\sqrt{2}\), (x=50), and \(x^{\frac{3}{2}}=x\sqrt{x}=50\cdot5\sqrt{2}=250\sqrt{2}\). In exams, write \(x^{\frac{3}{2}}\) as \(x\sqrt{x}\).
Step 2
Why this answer is correct
The correct answer is A. \(250\sqrt{2}\). From \(\sqrt{x}=5\sqrt{2}\), (x=50), and \(x^{\frac{3}{2}}=x\sqrt{x}=50\cdot5\sqrt{2}=250\sqrt{2}\). In exams, write \(x^{\frac{3}{2}}\) as \(x\sqrt{x}\).
Step 3
Exam Tip
\(\sqrt{x}=5\sqrt{2}\) से (x=50), और \(x^{\frac{3}{2}}=x\sqrt{x}=50\cdot5\sqrt{2}=250\sqrt{2}\)। परीक्षा में \(x^{\frac{3}{2}}\) को \(x\sqrt{x}\) लिखें।
We use (x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)). Thus (60=6\left\(x+\frac{1}{x}\right\)), so the value is (10).
Step 2
Why this answer is correct
The correct answer is C. (10). We use (x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)). Thus (60=6\left\(x+\frac{1}{x}\right\)), so the value is (10).
Step 3
Exam Tip
(x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)) है। इसलिए (60=6\left\(x+\frac{1}{x}\right\)) और मान (10) है।
We get (\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{\frac{1}{3}}=\frac{5x^{-3}}{4y^{4}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{4x^{3}y^{4}}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4x^{3}y^{4}}{5}\). We get (\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{\frac{1}{3}}=\frac{5x^{-3}}{4y^{4}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{4x^{3}y^{4}}{5}\).
Step 3
Exam Tip
(\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{\frac{1}{3}}=\frac{5x^{-3}}{4y^{4}})। \(-\frac{1}{3}\) घात लेने पर व्युत्क्रम \(\frac{4x^{3}y^{4}}{5}\) मिलता है।
Here \(\frac{1}{s}=\sqrt{17}-4\), so \(s-\frac{1}{s}=8\) and \(s+\frac{1}{s}=2\sqrt{17}\). Thus \(s^{2}-\frac{1}{s^{2}}=16\sqrt{17}\).
Step 2
Why this answer is correct
The correct answer is A. \(16\sqrt{17}\). Here \(\frac{1}{s}=\sqrt{17}-4\), so \(s-\frac{1}{s}=8\) and \(s+\frac{1}{s}=2\sqrt{17}\). Thus \(s^{2}-\frac{1}{s^{2}}=16\sqrt{17}\).
Step 3
Exam Tip
\(\frac{1}{s}=\sqrt{17}-4\), इसलिए \(s-\frac{1}{s}=8\) और \(s+\frac{1}{s}=2\sqrt{17}\)। अतः \(s^{2}-\frac{1}{s^{2}}=16\sqrt{17}\)।
Since (24^{3}=\(2^{3}\cdot3\)^{3}=2^{9}\cdot3^{3}), division leaves \(2^{3}\cdot3=24\), so the correct value is not among the options.
Step 2
Why this answer is correct
The correct answer is B. (6). Since (24^{3}=\(2^{3}\cdot3\)^{3}=2^{9}\cdot3^{3}), division leaves \(2^{3}\cdot3=24\), so the correct value is not among the options.
Step 3
Exam Tip
(24^{3}=\(2^{3}\cdot3\)^{3}=2^{9}\cdot3^{3})। भाग देने पर \(2^{3}\cdot3=24\) मिलता है, इसलिए विकल्पों में सही मान नहीं है।
Since \(1024=2^{10}\), \(16^{x}=2^{4x}\) gives \(x=\frac{5}{2}\), and \(32^{y}=2^{5y}\) gives (y=2). Hence the sum is \(\frac{9}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9}{2}\). Since \(1024=2^{10}\), \(16^{x}=2^{4x}\) gives \(x=\frac{5}{2}\), and \(32^{y}=2^{5y}\) gives (y=2). Hence the sum is \(\frac{9}{2}\).
Step 3
Exam Tip
\(1024=2^{10}\), \(16^{x}=2^{4x}\) से \(x=\frac{5}{2}\), और \(32^{y}=2^{5y}\) से (y=2)। इसलिए योग \(\frac{9}{2}\) है।
Inside, \(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}=\frac{1}{9}r^{-6}s^{8}\). Raising to (-1) gives \(9r^{6}s^{-8}\).
Step 2
Why this answer is correct
The correct answer is A. \(9r^{6}s^{-8}\). Inside, \(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}=\frac{1}{9}r^{-6}s^{8}\). Raising to (-1) gives \(9r^{6}s^{-8}\).
Step 3
Exam Tip
अंदर \(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}=\frac{1}{9}r^{-6}s^{8}\) है। (-1) घात लेने पर \(9r^{6}s^{-8}\) मिलता है।
Since (x^{12}-4096=\(x^{6}\)^{2}-64^{2}=\(x^{6}-64\)\(x^{6}+64\)), cancelling the common factor gives \(x^{6}+64\).
Step 2
Why this answer is correct
The correct answer is B. \(x^{6}+64\). Since (x^{12}-4096=\(x^{6}\)^{2}-64^{2}=\(x^{6}-64\)\(x^{6}+64\)), cancelling the common factor gives \(x^{6}+64\).
Step 3
Exam Tip
(x^{12}-4096=\(x^{6}\)^{2}-64^{2}=\(x^{6}-64\)\(x^{6}+64\))। समान गुणनखंड कटने पर \(x^{6}+64\) मिलता है।
We have \(\sqrt[3]{343}=7\), \(\sqrt[3]{a^{15}}=a^{5}\), and \(\sqrt[3]{b^{12}}=b^{4}\). In exams, divide exponents by (3) under a cube root.
Step 2
Why this answer is correct
The correct answer is A. \(7a^{5}b^{4}\). We have \(\sqrt[3]{343}=7\), \(\sqrt[3]{a^{15}}=a^{5}\), and \(\sqrt[3]{b^{12}}=b^{4}\). In exams, divide exponents by (3) under a cube root.
Step 3
Exam Tip
\(\sqrt[3]{343}=7\), \(\sqrt[3]{a^{15}}=a^{5}\), और \(\sqrt[3]{b^{12}}=b^{4}\)। परीक्षा में घनमूल में घातों को (3) से भाग दें।
Here (125^{\frac{2}{3}}=(5)^{2}=25) and (25^{-\frac{3}{2}}=(5)^{-3}=\frac{1}{125}). The product is \(\frac{1}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{5}\). Here (125^{\frac{2}{3}}=(5)^{2}=25) and (25^{-\frac{3}{2}}=(5)^{-3}=\frac{1}{125}). The product is \(\frac{1}{5}\).
Step 3
Exam Tip
(125^{\frac{2}{3}}=(5)^{2}=25) और (25^{-\frac{3}{2}}=(5)^{-3}=\frac{1}{125})। गुणनफल \(\frac{1}{5}\) है।
Inside, \(\frac{x^{-4}y^{5}}{z^{-2}}=x^{-4}y^{5}z^{2}\), so its reciprocal is \(x^{4}y^{-5}z^{-2}\). Multiplying by \(\frac{y^{3}}{x^{2}z^{4}}\) gives \(\frac{x^{2}}{y^{2}z^{6}}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{x^{2}}{y^{2}z^{2}}\). Inside, \(\frac{x^{-4}y^{5}}{z^{-2}}=x^{-4}y^{5}z^{2}\), so its reciprocal is \(x^{4}y^{-5}z^{-2}\). Multiplying by \(\frac{y^{3}}{x^{2}z^{4}}\) gives \(\frac{x^{2}}{y^{2}z^{6}}\).
Step 3
Exam Tip
अंदर \(\frac{x^{-4}y^{5}}{z^{-2}}=x^{-4}y^{5}z^{2}\), इसलिए उल्टा \(x^{4}y^{-5}z^{-2}\) है। \(\frac{y^{3}}{x^{2}z^{4}}\) से गुणा करने पर \(\frac{x^{2}}{y^{2}z^{6}}\) मिलता है।
Here \(\sqrt{300}=10\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\), and \(\sqrt{108}=6\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value is (12).
Step 2
Why this answer is correct
The correct answer is C. (12). Here \(\sqrt{300}=10\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\), and \(\sqrt{108}=6\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value is (12).
Step 3
Exam Tip
\(\sqrt{300}=10\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\), और \(\sqrt{108}=6\sqrt{3}\)। अंश \(12\sqrt{3}\) है, इसलिए मान (12) है।
We get (a=4), and \(9^{b}=3^{2b}=3^{6}\) gives (b=3). Thus \(a^{b}-b^{a}=4^{3}-3^{4}=64-81=-17\), which is not among the options.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{37}{8}\). We get (a=4), and \(9^{b}=3^{2b}=3^{6}\) gives (b=3). Thus \(a^{b}-b^{a}=4^{3}-3^{4}=64-81=-17\), which is not among the options.
Step 3
Exam Tip
(a=4) और \(9^{b}=3^{2b}=3^{6}\) से (b=3) है। इसलिए \(a^{b}-b^{a}=4^{3}-3^{4}=64-81=-17\), अतः विकल्पों में यह मान नहीं है।
Here (\(4\sqrt{3}\)^{2}=48), (\(3\sqrt{5}\)^{2}=45), and the middle term is \(24\sqrt{15}\). Therefore, the expansion is \(93-24\sqrt{15}\).
Step 2
Why this answer is correct
The correct answer is A. \(93-24\sqrt{15}\). Here (\(4\sqrt{3}\)^{2}=48), (\(3\sqrt{5}\)^{2}=45), and the middle term is \(24\sqrt{15}\). Therefore, the expansion is \(93-24\sqrt{15}\).
Step 3
Exam Tip
(\(4\sqrt{3}\)^{2}=48), (\(3\sqrt{5}\)^{2}=45), और मध्य पद \(24\sqrt{15}\) है। इसलिए विस्तार \(93-24\sqrt{15}\) है।
Since (\left\(\frac{25}{49}\right\)^{\frac{1}{2}}=\frac{5}{7}), (\left\(\frac{25}{49}\right\)^{-\frac{3}{2}}=\left\(\frac{5}{7}\right\)^{-3}=\frac{343}{125}). In exams, take the square root first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{343}{125}\). Since (\left\(\frac{25}{49}\right\)^{\frac{1}{2}}=\frac{5}{7}), (\left\(\frac{25}{49}\right\)^{-\frac{3}{2}}=\left\(\frac{5}{7}\right\)^{-3}=\frac{343}{125}). In exams, take the square root first.
Step 3
Exam Tip
(\left\(\frac{25}{49}\right\)^{\frac{1}{2}}=\frac{5}{7}), इसलिए (\left\(\frac{25}{49}\right\)^{-\frac{3}{2}}=\left\(\frac{5}{7}\right\)^{-3}=\frac{343}{125})। परीक्षा में पहले वर्गमूल निकालें।
The product of denominators is (26-25=1), and the numerator is (\(\sqrt{26}+5\)+\(\sqrt{26}-5\)=2\sqrt{26}). In exams, add conjugate fractions together.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{26}\). The product of denominators is (26-25=1), and the numerator is (\(\sqrt{26}+5\)+\(\sqrt{26}-5\)=2\sqrt{26}). In exams, add conjugate fractions together.
Step 3
Exam Tip
हरों का गुणनफल (26-25=1) है और अंश (\(\sqrt{26}+5\)+\(\sqrt{26}-5\)=2\sqrt{26}) है। परीक्षा में संयुग्म भिन्नों को साथ जोड़ें।
We use (x^{10}-1024=\(x^{5}\)^{2}-32^{2}=\(x^{5}-32\)\(x^{5}+32\)). Cancelling the common factor leaves \(x^{5}+32\).
Step 2
Why this answer is correct
The correct answer is B. \(x^{5}+32\). We use (x^{10}-1024=\(x^{5}\)^{2}-32^{2}=\(x^{5}-32\)\(x^{5}+32\)). Cancelling the common factor leaves \(x^{5}+32\).
Step 3
Exam Tip
(x^{10}-1024=\(x^{5}\)^{2}-32^{2}=\(x^{5}-32\)\(x^{5}+32\))। समान गुणनखंड कटने पर \(x^{5}+32\) बचता है।
Here (25^{\frac{3}{2}}=\(5^{2}\)^{\frac{3}{2}}=5^{3}) and (125^{-\frac{2}{3}}=\(5^{3}\)^{-\frac{2}{3}}=5^{-2}). The product is (5).
Step 2
Why this answer is correct
The correct answer is B. (5). Here (25^{\frac{3}{2}}=\(5^{2}\)^{\frac{3}{2}}=5^{3}) and (125^{-\frac{2}{3}}=\(5^{3}\)^{-\frac{2}{3}}=5^{-2}). The product is (5).
Step 3
Exam Tip
(25^{\frac{3}{2}}=\(5^{2}\)^{\frac{3}{2}}=5^{3}) और (125^{-\frac{2}{3}}=\(5^{3}\)^{-\frac{2}{3}}=5^{-2})। गुणनफल (5) है।
Inside, \(\frac{8x^{-3}y^{2}}{2x^{5}y^{-4}}=4x^{-8}y^{6}\), and its square is \(16x^{-16}y^{12}\). Multiplying by \(\frac{x^{16}}{16y^{12}}\) gives (1).
Step 2
Why this answer is correct
The correct answer is A. (1). Inside, \(\frac{8x^{-3}y^{2}}{2x^{5}y^{-4}}=4x^{-8}y^{6}\), and its square is \(16x^{-16}y^{12}\). Multiplying by \(\frac{x^{16}}{16y^{12}}\) gives (1).
Step 3
Exam Tip
अंदर \(\frac{8x^{-3}y^{2}}{2x^{5}y^{-4}}=4x^{-8}y^{6}\), इसका वर्ग \(16x^{-16}y^{12}\) है। फिर \(\frac{x^{16}}{16y^{12}}\) से गुणा करने पर (1) मिलता है।
Because (\(3+\sqrt{10}\)^{2}=9+10+6\sqrt{10}=19+6\sqrt{10}), \(\sqrt{A}=3+\sqrt{10}\). In exams, identify perfect-square surd forms.
Step 2
Why this answer is correct
The correct answer is A. \(3+\sqrt{10}\). Because (\(3+\sqrt{10}\)^{2}=9+10+6\sqrt{10}=19+6\sqrt{10}), \(\sqrt{A}=3+\sqrt{10}\). In exams, identify perfect-square surd forms.
Step 3
Exam Tip
क्योंकि (\(3+\sqrt{10}\)^{2}=9+10+6\sqrt{10}=19+6\sqrt{10}), इसलिए \(\sqrt{A}=3+\sqrt{10}\)। परीक्षा में पूर्ण वर्ग करणी पहचानें।
Let \(A=x^{-2}\) and \(B=y^{-2}\). Then \(\frac{A^{2}-B^{2}}{A-B}=A+B\), so the answer is \(x^{-2}+y^{-2}=\frac{x^{2}+y^{2}}{x^{2}y^{2}}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{x^{2}+y^{2}}{x^{2}y^{2}}\). Let \(A=x^{-2}\) and \(B=y^{-2}\). Then \(\frac{A^{2}-B^{2}}{A-B}=A+B\), so the answer is \(x^{-2}+y^{-2}=\frac{x^{2}+y^{2}}{x^{2}y^{2}}\).
Step 3
Exam Tip
मान लें \(A=x^{-2}\) और \(B=y^{-2}\), तो \(\frac{A^{2}-B^{2}}{A-B}=A+B\)। इसलिए उत्तर \(x^{-2}+y^{-2}=\frac{x^{2}+y^{2}}{x^{2}y^{2}}\) है।
We have \(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), \(\sqrt{98}=7\sqrt{2}\), and \(\sqrt{72}=6\sqrt{2}\). The total is \(4\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is C. \(4\sqrt{2}\). We have \(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), \(\sqrt{98}=7\sqrt{2}\), and \(\sqrt{72}=6\sqrt{2}\). The total is \(4\sqrt{2}\).
Step 3
Exam Tip
\(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), \(\sqrt{98}=7\sqrt{2}\), और \(\sqrt{72}=6\sqrt{2}\)। कुल \(4\sqrt{2}\) मिलता है।
Here \(169^{-1}=13^{-2}\) and \(2197^{-1}=13^{-3}\), so \(\frac{13^{4}\cdot13^{-2}}{13^{-3}}=13^{5}\). In exams, division by a negative power adds the exponent.
Step 2
Why this answer is correct
The correct answer is B. \(13^{5}\). Here \(169^{-1}=13^{-2}\) and \(2197^{-1}=13^{-3}\), so \(\frac{13^{4}\cdot13^{-2}}{13^{-3}}=13^{5}\). In exams, division by a negative power adds the exponent.
Step 3
Exam Tip
\(169^{-1}=13^{-2}\) और \(2197^{-1}=13^{-3}\), इसलिए \(\frac{13^{4}\cdot13^{-2}}{13^{-3}}=13^{5}\)। परीक्षा में ऋणात्मक घात से भाग करते समय घात जुड़ती है।
We use (\left\(x+\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}+2). Thus \(49=x^{2}+\frac{1}{x^{2}}+2\), so the value is (47).
Step 2
Why this answer is correct
The correct answer is B. (47). We use (\left\(x+\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}+2). Thus \(49=x^{2}+\frac{1}{x^{2}}+2\), so the value is (47).
Step 3
Exam Tip
(\left\(x+\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}+2) होता है। इसलिए \(49=x^{2}+\frac{1}{x^{2}}+2\) और मान (47) है।
Since (\left\(\frac{81}{256}\right\)^{\frac{1}{4}}=\frac{3}{4}), (\left\(\frac{81}{256}\right\)^{-\frac{3}{4}}=\left\(\frac{3}{4}\right\)^{-3}=\frac{64}{27}). In exams, take the fourth root first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{64}{27}\). Since (\left\(\frac{81}{256}\right\)^{\frac{1}{4}}=\frac{3}{4}), (\left\(\frac{81}{256}\right\)^{-\frac{3}{4}}=\left\(\frac{3}{4}\right\)^{-3}=\frac{64}{27}). In exams, take the fourth root first.
Step 3
Exam Tip
(\left\(\frac{81}{256}\right\)^{\frac{1}{4}}=\frac{3}{4}), इसलिए (\left\(\frac{81}{256}\right\)^{-\frac{3}{4}}=\left\(\frac{3}{4}\right\)^{-3}=\frac{64}{27})। परीक्षा में पहले चौथा मूल निकालें।
Here \(6^{x+1}-6^{x}=5\cdot6^{x}=900\), so \(6^{x}=180\), which is not a listed integral power. Therefore none of the listed options is correct.
Step 2
Why this answer is correct
The correct answer is B. (3). Here \(6^{x+1}-6^{x}=5\cdot6^{x}=900\), so \(6^{x}=180\), which is not a listed integral power. Therefore none of the listed options is correct.
Step 3
Exam Tip
\(6^{x+1}-6^{x}=6\cdot6^{x}-6^{x}=5\cdot6^{x}=900\), इसलिए \(6^{x}=180\) नहीं बनता। इसलिए दिए विकल्पों में कोई भी सही नहीं है।
The product of the denominators is (25-24=1), and the numerator becomes \(2\sqrt{24}\). In exams, first find the product of conjugate denominators.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{24}\). The product of the denominators is (25-24=1), and the numerator becomes \(2\sqrt{24}\). In exams, first find the product of conjugate denominators.
Step 3
Exam Tip
हरों का गुणनफल (25-24=1) है और अंश \(2\sqrt{24}\) बनता है। परीक्षा में संयुग्म हरों का गुणनफल पहले निकालें।
Here (u^{2}-v^{2}=(u-v)(u+v)=2\sqrt{8}\cdot2\sqrt{17}=8\sqrt{34}), and (uv=9). Hence the value is \(\frac{8\sqrt{34}}{9}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{8\sqrt{34}}{9}\). Here (u^{2}-v^{2}=(u-v)(u+v)=2\sqrt{8}\cdot2\sqrt{17}=8\sqrt{34}), and (uv=9). Hence the value is \(\frac{8\sqrt{34}}{9}\).
Step 3
Exam Tip
(u^{2}-v^{2}=(u-v)(u+v)=2\sqrt{8}\cdot2\sqrt{17}=8\sqrt{34}) और (uv=9) है। इसलिए मान \(\frac{8\sqrt{34}}{9}\) है।
Since \(25^{-2}=5^{-4}\) and \(125=5^{3}\), the total exponent is (9-4+3-4=4). In exams, convert all terms to the same base.
Step 2
Why this answer is correct
The correct answer is C. \(5^{4}\). Since \(25^{-2}=5^{-4}\) and \(125=5^{3}\), the total exponent is (9-4+3-4=4). In exams, convert all terms to the same base.
Step 3
Exam Tip
\(25^{-2}=5^{-4}\) और \(125=5^{3}\), इसलिए कुल घात (9-4+3-4=4) है। परीक्षा में सभी पदों को समान आधार में बदलें।
Here \(\frac{4x^{-2}}{x^{3}}=4x^{-5}\), so its reciprocal is \(\frac{x^{5}}{4}\), and multiplying by \(x^{-4}\) gives \(\frac{x}{4}\). In exams, simplify the bracket first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{x}{4}\). Here \(\frac{4x^{-2}}{x^{3}}=4x^{-5}\), so its reciprocal is \(\frac{x^{5}}{4}\), and multiplying by \(x^{-4}\) gives \(\frac{x}{4}\). In exams, simplify the bracket first.
Step 3
Exam Tip
\(\frac{4x^{-2}}{x^{3}}=4x^{-5}\), इसलिए व्युत्क्रम \(\frac{x^{5}}{4}\) है और \(x^{-4}\) से गुणा करने पर \(\frac{x}{4}\) मिलता है। परीक्षा में पहले कोष्ठक को सरल करें।
Here \(x^{2}=7+2\sqrt{10}\), so \(x^{3}=17\sqrt{2}+11\sqrt{5}\) and \(x^{3}-7x=10\sqrt{2}+4\sqrt{5}\). In exams, first find \(x^{2}\) and then multiply by (x).
Step 2
Why this answer is correct
The correct answer is A. \(10\sqrt{2}+4\sqrt{5}\). Here \(x^{2}=7+2\sqrt{10}\), so \(x^{3}=17\sqrt{2}+11\sqrt{5}\) and \(x^{3}-7x=10\sqrt{2}+4\sqrt{5}\). In exams, first find \(x^{2}\) and then multiply by (x).
Step 3
Exam Tip
\(x^{2}=7+2\sqrt{10}\), इसलिए \(x^{3}=17\sqrt{2}+11\sqrt{5}\) और \(x^{3}-7x=10\sqrt{2}+4\sqrt{5}\)। परीक्षा में पहले \(x^{2}\) निकालकर फिर (x) से गुणा करें।
Here \(\sqrt{192}=8\sqrt{3}\), \(2\sqrt{48}=8\sqrt{3}\), and \(3\sqrt{12}=6\sqrt{3}\). The numerator is \(6\sqrt{3}\), so the value is (6).
Step 2
Why this answer is correct
The correct answer is C. (12). Here \(\sqrt{192}=8\sqrt{3}\), \(2\sqrt{48}=8\sqrt{3}\), and \(3\sqrt{12}=6\sqrt{3}\). The numerator is \(6\sqrt{3}\), so the value is (6).
Step 3
Exam Tip
\(\sqrt{192}=8\sqrt{3}\), \(2\sqrt{48}=8\sqrt{3}\), और \(3\sqrt{12}=6\sqrt{3}\)। अंश \(6\sqrt{3}\) है, इसलिए मान (6) है।
Here (\left\(\frac{4}{7}\right\)^{-2}=\frac{49}{16}) and (\left\(\frac{7}{4}\right\)^{-2}=\frac{16}{49}). The sum is \(\frac{2401+256}{784}=\frac{2657}{784}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{2657}{784}\). Here (\left\(\frac{4}{7}\right\)^{-2}=\frac{49}{16}) and (\left\(\frac{7}{4}\right\)^{-2}=\frac{16}{49}). The sum is \(\frac{2401+256}{784}=\frac{2657}{784}\).
Step 3
Exam Tip
(\left\(\frac{4}{7}\right\)^{-2}=\frac{49}{16}) और (\left\(\frac{7}{4}\right\)^{-2}=\frac{16}{49})। योग \(\frac{2401+256}{784}=\frac{2657}{784}\) है।
From \(\sqrt{x}=4\sqrt{3}\), (x=48), and \(x^{\frac{3}{2}}=x\sqrt{x}=48\cdot4\sqrt{3}=192\sqrt{3}\). In exams, write \(x^{\frac{3}{2}}\) as \(x\sqrt{x}\).
Step 2
Why this answer is correct
The correct answer is A. \(192\sqrt{3}\). From \(\sqrt{x}=4\sqrt{3}\), (x=48), and \(x^{\frac{3}{2}}=x\sqrt{x}=48\cdot4\sqrt{3}=192\sqrt{3}\). In exams, write \(x^{\frac{3}{2}}\) as \(x\sqrt{x}\).
Step 3
Exam Tip
\(\sqrt{x}=4\sqrt{3}\) से (x=48), और \(x^{\frac{3}{2}}=x\sqrt{x}=48\cdot4\sqrt{3}=192\sqrt{3}\)। परीक्षा में \(x^{\frac{3}{2}}\) को \(x\sqrt{x}\) लिखें।
We use (x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)). Thus (40=5\left\(x+\frac{1}{x}\right\)), so the value is (8).
Step 2
Why this answer is correct
The correct answer is C. (8). We use (x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)). Thus (40=5\left\(x+\frac{1}{x}\right\)), so the value is (8).
Step 3
Exam Tip
(x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)) है। इसलिए (40=5\left\(x+\frac{1}{x}\right\)), और मान (8) है।
We get (\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{\frac{1}{3}}=\frac{4x^{-2}}{3y^{3}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{3x^{2}y^{3}}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3x^{2}y^{3}}{4}\). We get (\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{\frac{1}{3}}=\frac{4x^{-2}}{3y^{3}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{3x^{2}y^{3}}{4}\).
Step 3
Exam Tip
(\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{\frac{1}{3}}=\frac{4x^{-2}}{3y^{3}})। \(-\frac{1}{3}\) घात लेने पर व्युत्क्रम \(\frac{3x^{2}y^{3}}{4}\) मिलता है।
Here \(\frac{1}{s}=\sqrt{10}-3\), so \(s-\frac{1}{s}=6\) and \(s+\frac{1}{s}=2\sqrt{10}\). Thus \(s^{2}-\frac{1}{s^{2}}=12\sqrt{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(12\sqrt{10}\). Here \(\frac{1}{s}=\sqrt{10}-3\), so \(s-\frac{1}{s}=6\) and \(s+\frac{1}{s}=2\sqrt{10}\). Thus \(s^{2}-\frac{1}{s^{2}}=12\sqrt{10}\).
Step 3
Exam Tip
\(\frac{1}{s}=\sqrt{10}-3\), इसलिए \(s-\frac{1}{s}=6\) और \(s+\frac{1}{s}=2\sqrt{10}\)। अतः \(s^{2}-\frac{1}{s^{2}}=12\sqrt{10}\)।
Inside, \(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}=\frac{1}{7}r^{-5}s^{6}\). Raising to (-1) gives \(7r^{5}s^{-6}\).
Step 2
Why this answer is correct
The correct answer is A. \(7r^{5}s^{-6}\). Inside, \(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}=\frac{1}{7}r^{-5}s^{6}\). Raising to (-1) gives \(7r^{5}s^{-6}\).
Step 3
Exam Tip
अंदर \(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}=\frac{1}{7}r^{-5}s^{6}\) है। (-1) घात लेने पर \(7r^{5}s^{-6}\) मिलता है।
Here \(3^{-2}+3^{-4}=\frac{1}{9}+\frac{1}{81}=\frac{10}{81}\), and \(3^{-3}=\frac{1}{27}\). Division gives \(\frac{10}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{10}{3}\). Here \(3^{-2}+3^{-4}=\frac{1}{9}+\frac{1}{81}=\frac{10}{81}\), and \(3^{-3}=\frac{1}{27}\). Division gives \(\frac{10}{3}\).
Step 3
Exam Tip
\(3^{-2}+3^{-4}=\frac{1}{9}+\frac{1}{81}=\frac{10}{81}\) और \(3^{-3}=\frac{1}{27}\)। भाग देने पर \(\frac{10}{3}\) मिलता है।
We have \(\sqrt[3]{216}=6\), \(\sqrt[3]{a^{12}}=a^{4}\), and \(\sqrt[3]{b^{9}}=b^{3}\). In exams, divide exponents by (3) under a cube root.
Step 2
Why this answer is correct
The correct answer is A. \(6a^{4}b^{3}\). We have \(\sqrt[3]{216}=6\), \(\sqrt[3]{a^{12}}=a^{4}\), and \(\sqrt[3]{b^{9}}=b^{3}\). In exams, divide exponents by (3) under a cube root.
Step 3
Exam Tip
\(\sqrt[3]{216}=6\), \(\sqrt[3]{a^{12}}=a^{4}\), और \(\sqrt[3]{b^{9}}=b^{3}\)। परीक्षा में घनमूल में घातों को (3) से भाग दें।
Inside, \(\frac{x^{-2}y^{4}}{z^{-3}}=x^{-2}y^{4}z^{3}\), so its reciprocal is \(x^{2}y^{-4}z^{-3}\). Multiplying by \(\frac{y^{2}}{x^{3}z^{2}}\) gives \(\frac{1}{xy^{2}z^{5}}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{z}{xy^{2}}\). Inside, \(\frac{x^{-2}y^{4}}{z^{-3}}=x^{-2}y^{4}z^{3}\), so its reciprocal is \(x^{2}y^{-4}z^{-3}\). Multiplying by \(\frac{y^{2}}{x^{3}z^{2}}\) gives \(\frac{1}{xy^{2}z^{5}}\).
Step 3
Exam Tip
अंदर \(\frac{x^{-2}y^{4}}{z^{-3}}=x^{-2}y^{4}z^{3}\), इसलिए उल्टा \(x^{2}y^{-4}z^{-3}\) है। \(\frac{y^{2}}{x^{3}z^{2}}\) से गुणा करने पर \(\frac{1}{xy^{2}z^{5}}\) मिलता है।
Here \(\sqrt{108}=6\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), and \(\sqrt{12}=2\sqrt{3}\). The numerator is \(9\sqrt{3}\), so the value is (9).
Step 2
Why this answer is correct
The correct answer is C. (9). Here \(\sqrt{108}=6\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), and \(\sqrt{12}=2\sqrt{3}\). The numerator is \(9\sqrt{3}\), so the value is (9).
Step 3
Exam Tip
\(\sqrt{108}=6\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), और \(\sqrt{12}=2\sqrt{3}\)। अंश \(9\sqrt{3}\) है, इसलिए मान (9) है।
