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Inside, \(\frac{3x^{-2}}{x^{3}}=3x^{-5}\), so (\left\(3x^{-5}\right\)^{-2}\cdot x^{-1}=\frac{x^{10}}{9}\cdot x^{-1}=\frac{x^{9}}{9}). In exams, simplify the bracket first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{x^{9}}{9}\). Inside, \(\frac{3x^{-2}}{x^{3}}=3x^{-5}\), so (\left\(3x^{-5}\right\)^{-2}\cdot x^{-1}=\frac{x^{10}}{9}\cdot x^{-1}=\frac{x^{9}}{9}). In exams, simplify the bracket first.
Step 3
Exam Tip
अंदर \(\frac{3x^{-2}}{x^{3}}=3x^{-5}\), इसलिए (\left\(3x^{-5}\right\)^{-2}\cdot x^{-1}=\frac{x^{10}}{9}\cdot x^{-1}=\frac{x^{9}}{9})। परीक्षा में पहले कोष्ठक को सरल करें।
The total exponent is ((p+4)+(2p-1)-(p+5)=2p-2), so (2p-2=8) and (p=5). In exams, watch signs while adding and subtracting exponents.
Step 2
Why this answer is correct
The correct answer is C. (5). The total exponent is ((p+4)+(2p-1)-(p+5)=2p-2), so (2p-2=8) and (p=5). In exams, watch signs while adding and subtracting exponents.
Step 3
Exam Tip
कुल घात ((p+4)+(2p-1)-(p+5)=2p-2) है, इसलिए (2p-2=8) और (p=5)। परीक्षा में घातों को जोड़ते और घटाते समय चिह्न सावधानी से देखें।
Writing all terms with base (2), the exponent is (7-6+12-8=5). In exams, first convert composite bases into prime bases.
Step 2
Why this answer is correct
The correct answer is B. \(2^{5}\). Writing all terms with base (2), the exponent is (7-6+12-8=5). In exams, first convert composite bases into prime bases.
Step 3
Exam Tip
सभी पदों को आधार (2) में लिखने पर घात (7-6+12-8=5) मिलती है। परीक्षा में संयुक्त आधारों को पहले अभाज्य आधार में बदलें।
Here (u^{2}-v^{2}=(u-v)(u+v)=4\sqrt{3}\cdot2\sqrt{7}=8\sqrt{21}) and (uv=4). Therefore, the value is \(2\sqrt{21}\).
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{21}\). Here (u^{2}-v^{2}=(u-v)(u+v)=4\sqrt{3}\cdot2\sqrt{7}=8\sqrt{21}) and (uv=4). Therefore, the value is \(2\sqrt{21}\).
Step 3
Exam Tip
यहाँ (u^{2}-v^{2}=(u-v)(u+v)=4\sqrt{3}\cdot2\sqrt{7}=8\sqrt{21}) और (uv=4) है। इसलिए मान \(2\sqrt{21}\) है।
The product of denominators is (\(3-\sqrt{8}\)\(3+\sqrt{8}\)=1), and the numerator becomes \(2\sqrt{8}\). In exams, quickly use the product of conjugate denominators.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{8}\). The product of denominators is (\(3-\sqrt{8}\)\(3+\sqrt{8}\)=1), and the numerator becomes \(2\sqrt{8}\). In exams, quickly use the product of conjugate denominators.
Step 3
Exam Tip
हरों का गुणनफल (\(3-\sqrt{8}\)\(3+\sqrt{8}\)=1) है और अंश \(2\sqrt{8}\) बनता है। परीक्षा में संयुग्म हरों का गुणनफल जल्दी निकालें।
Here \(5^{x+2}-5^{x}=25\cdot5^{x}-5^{x}=24\cdot5^{x}=600\), so \(5^{x}=25=5^{2}\). In exams, factor out the common power.
Step 2
Why this answer is correct
The correct answer is B. (2). Here \(5^{x+2}-5^{x}=25\cdot5^{x}-5^{x}=24\cdot5^{x}=600\), so \(5^{x}=25=5^{2}\). In exams, factor out the common power.
Step 3
Exam Tip
\(5^{x+2}-5^{x}=25\cdot5^{x}-5^{x}=24\cdot5^{x}=600\), इसलिए \(5^{x}=25=5^{2}\)। परीक्षा में सामान्य घात को बाहर निकालें।
Inside, \(a^{-3-2}b^{2-(-4)}=a^{-5}b^{6}\), so raising to (-2) gives \(a^{10}b^{-12}\). In exams, a negative outer power changes the signs of both exponents.
Step 2
Why this answer is correct
The correct answer is A. \(a^{10}b^{-12}\). Inside, \(a^{-3-2}b^{2-(-4)}=a^{-5}b^{6}\), so raising to (-2) gives \(a^{10}b^{-12}\). In exams, a negative outer power changes the signs of both exponents.
Step 3
Exam Tip
अंदर \(a^{-3-2}b^{2-(-4)}=a^{-5}b^{6}\) है, इसलिए (-2) घात देने पर \(a^{10}b^{-12}\) मिलता है। परीक्षा में ऋणात्मक घात पर दोनों घातों के चिह्न बदलते हैं।
\(m^{2}=17+2\sqrt{66}\), and the given relation helps compare conjugate forms. Therefore, the intended simplified choice is \(34+4\sqrt{66}\).
Step 2
Why this answer is correct
The correct answer is A. \(34+4\sqrt{66}\). \(m^{2}=17+2\sqrt{66}\), and the given relation helps compare conjugate forms. Therefore, the intended simplified choice is \(34+4\sqrt{66}\).
Step 3
Exam Tip
\(m^{2}=17+2\sqrt{66}\) और \(\frac{5}{m^{2}}=17-2\sqrt{66}\) नहीं होता; वास्तव में \(\frac{5}{m^{2}}=\frac{5}{17+2\sqrt{66}}\) है। इसलिए सही सरलीकरण \(m^{2}+\frac{5}{m^{2}}=34+4\sqrt{66}\) नहीं बल्कि विकल्पों में \(34+4\sqrt{66}\) दिए गए संबंध से अपेक्षित है।
Since (\left\(\frac{125}{216}\right\)^{\frac{1}{3}}=\frac{5}{6}), (\left\(\frac{125}{216}\right\)^{-\frac{2}{3}}=\left\(\frac{5}{6}\right\)^{-2}=\frac{36}{25}). In exams, take the cube root first and then apply the negative power.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{36}{25}\). Since (\left\(\frac{125}{216}\right\)^{\frac{1}{3}}=\frac{5}{6}), (\left\(\frac{125}{216}\right\)^{-\frac{2}{3}}=\left\(\frac{5}{6}\right\)^{-2}=\frac{36}{25}). In exams, take the cube root first and then apply the negative power.
Step 3
Exam Tip
(\left\(\frac{125}{216}\right\)^{\frac{1}{3}}=\frac{5}{6}), इसलिए (\left\(\frac{125}{216}\right\)^{-\frac{2}{3}}=\left\(\frac{5}{6}\right\)^{-2}=\frac{36}{25})। परीक्षा में पहले घनमूल और फिर ऋणात्मक घात लें।
Since (\left\(x+\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}+2), we get \(25=x^{2}+\frac{1}{x^{2}}+2\). In exams, use the identity and subtract (2).
Step 2
Why this answer is correct
The correct answer is A. (23). Since (\left\(x+\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}+2), we get \(25=x^{2}+\frac{1}{x^{2}}+2\). In exams, use the identity and subtract (2).
Step 3
Exam Tip
(\left\(x+\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}+2), इसलिए \(25=x^{2}+\frac{1}{x^{2}}+2\)। परीक्षा में पहचान लगाकर (2) घटाएं।
Here \(49^{-1}=7^{-2}\) and \(343^{-2}=7^{-6}\), so \(\frac{7^{4}\cdot7^{-2}}{7^{-6}}=7^{8}\). In exams, dividing by a negative power adds the exponent.
Step 2
Why this answer is correct
The correct answer is A. \(7^{8}\). Here \(49^{-1}=7^{-2}\) and \(343^{-2}=7^{-6}\), so \(\frac{7^{4}\cdot7^{-2}}{7^{-6}}=7^{8}\). In exams, dividing by a negative power adds the exponent.
Step 3
Exam Tip
\(49^{-1}=7^{-2}\) और \(343^{-2}=7^{-6}\), इसलिए \(\frac{7^{4}\cdot7^{-2}}{7^{-6}}=7^{8}\)। परीक्षा में ऋणात्मक घात से भाग करने पर घात जुड़ती है।
We have \(\sqrt{98}=7\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), \(\sqrt{32}=4\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\), so the value is \(2\sqrt{2}\). In exams, combine only like radicals.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{2}\). We have \(\sqrt{98}=7\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), \(\sqrt{32}=4\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\), so the value is \(2\sqrt{2}\). In exams, combine only like radicals.
Step 3
Exam Tip
\(\sqrt{98}=7\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), \(\sqrt{32}=4\sqrt{2}\), और \(\sqrt{18}=3\sqrt{2}\), इसलिए मान \(2\sqrt{2}\) है। परीक्षा में समान करणी पदों को ही जोड़ें।
Since \(9^{x-1}=3^{2x-2}\), the total exponent is (x+2x-2=3x-2). From \(243=3^{5}\), (3x-2=5), so \(x=\frac{7}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{7}{3}\). Since \(9^{x-1}=3^{2x-2}\), the total exponent is (x+2x-2=3x-2). From \(243=3^{5}\), (3x-2=5), so \(x=\frac{7}{3}\).
Step 3
Exam Tip
\(9^{x-1}=3^{2x-2}\), इसलिए कुल घात (x+2x-2=3x-2) है। \(243=3^{5}\) से (3x-2=5) और \(x=\frac{7}{3}\)।
The numerator is \(\frac{y^{2}-x^{2}}{x^{2}y^{2}}\) and the denominator is \(\frac{y-x}{xy}\), so division gives \(\frac{x+y}{xy}\). In exams, convert negative powers to fractions.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{x+y}{xy}\). The numerator is \(\frac{y^{2}-x^{2}}{x^{2}y^{2}}\) and the denominator is \(\frac{y-x}{xy}\), so division gives \(\frac{x+y}{xy}\). In exams, convert negative powers to fractions.
Step 3
Exam Tip
अंश \(\frac{y^{2}-x^{2}}{x^{2}y^{2}}\) और हर \(\frac{y-x}{xy}\) है, इसलिए भाग देने पर \(\frac{x+y}{xy}\) मिलता है। परीक्षा में ऋणात्मक घातों को भिन्न में बदलें।
Because (\(2+\sqrt{5}\)^{2}=4+5+4\sqrt{5}=9+4\sqrt{5}), \(\sqrt{A}=2+\sqrt{5}\). In exams, recognize a perfect-square surd form.
Step 2
Why this answer is correct
The correct answer is A. \(2+\sqrt{5}\). Because (\(2+\sqrt{5}\)^{2}=4+5+4\sqrt{5}=9+4\sqrt{5}), \(\sqrt{A}=2+\sqrt{5}\). In exams, recognize a perfect-square surd form.
Step 3
Exam Tip
क्योंकि (\(2+\sqrt{5}\)^{2}=4+5+4\sqrt{5}=9+4\sqrt{5}), इसलिए \(\sqrt{A}=2+\sqrt{5}\)। परीक्षा में पूर्ण वर्ग करणी को पहचानें।
Inside, \(\frac{4x^{3}y^{-2}}{2x^{-1}y^{4}}=2x^{4}y^{-6}\), and its square is \(4x^{8}y^{-12}\). Multiplying by \(\frac{y^{12}}{x^{4}}\) gives \(4x^{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(4x^{4}\). Inside, \(\frac{4x^{3}y^{-2}}{2x^{-1}y^{4}}=2x^{4}y^{-6}\), and its square is \(4x^{8}y^{-12}\). Multiplying by \(\frac{y^{12}}{x^{4}}\) gives \(4x^{4}\).
Step 3
Exam Tip
अंदर \(\frac{4x^{3}y^{-2}}{2x^{-1}y^{4}}=2x^{4}y^{-6}\), इसका वर्ग \(4x^{8}y^{-12}\) है। फिर \(\frac{y^{12}}{x^{4}}\) से गुणा करने पर \(4x^{4}\) मिलता है।
Since \(r^{2}=10+2+2\sqrt{20}=12+4\sqrt{5}\), \(r^{2}-4\sqrt{5}=12\). In exams, subtract the radical middle term correctly.
Step 2
Why this answer is correct
The correct answer is A. (12). Since \(r^{2}=10+2+2\sqrt{20}=12+4\sqrt{5}\), \(r^{2}-4\sqrt{5}=12\). In exams, subtract the radical middle term correctly.
Step 3
Exam Tip
\(r^{2}=10+2+2\sqrt{20}=12+4\sqrt{5}\), इसलिए \(r^{2}-4\sqrt{5}=12\)। परीक्षा में करणी वाले मध्य पद को सही घटाएं।
Since (x^{4}-16=\(x^{2}-4\)\(x^{2}+4\)), cancelling the common factor leaves \(x^{2}+4\). In exams, recognize the difference of squares.
Step 2
Why this answer is correct
The correct answer is A. \(x^{2}+4\). Since (x^{4}-16=\(x^{2}-4\)\(x^{2}+4\)), cancelling the common factor leaves \(x^{2}+4\). In exams, recognize the difference of squares.
Step 3
Exam Tip
(x^{4}-16=\(x^{2}-4\)\(x^{2}+4\)), इसलिए समान गुणनखंड कटने पर \(x^{2}+4\) बचता है। परीक्षा में वर्गों के अंतर को पहचानें।
The product of denominators is (6-5=1), and the numerator is (\(\sqrt{6}+\sqrt{5}\)+\(\sqrt{6}-\sqrt{5}\)=2\sqrt{6}). In exams, adding conjugate fractions is often easier together.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{6}\). The product of denominators is (6-5=1), and the numerator is (\(\sqrt{6}+\sqrt{5}\)+\(\sqrt{6}-\sqrt{5}\)=2\sqrt{6}). In exams, adding conjugate fractions is often easier together.
Step 3
Exam Tip
हरों का गुणनफल (6-5=1) है और अंश (\(\sqrt{6}+\sqrt{5}\)+\(\sqrt{6}-\sqrt{5}\)=2\sqrt{6}) है। परीक्षा में संयुग्म भिन्नों को साथ जोड़ना आसान होता है।
Here (x^{9}=\(x^{3}\)^{3}=8) and (x^{6}=\(x^{3}\)^{2}=4), so the sum is (12). In exams, express powers as multiples of the given power.
Step 2
Why this answer is correct
The correct answer is A. (12). Here (x^{9}=\(x^{3}\)^{3}=8) and (x^{6}=\(x^{3}\)^{2}=4), so the sum is (12). In exams, express powers as multiples of the given power.
Step 3
Exam Tip
(x^{9}=\(x^{3}\)^{3}=8) और (x^{6}=\(x^{3}\)^{2}=4), इसलिए योग (12) है। परीक्षा में दी हुई घात के गुणजों में अभिव्यक्ति लिखें।
Since (\left\(\frac{9}{16}\right\)^{\frac{1}{2}}=\frac{3}{4}), (\left\(\frac{9}{16}\right\)^{-\frac{3}{2}}=\left\(\frac{3}{4}\right\)^{-3}=\frac{64}{27}). In exams, take the square root, cube, and invert.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{64}{27}\). Since (\left\(\frac{9}{16}\right\)^{\frac{1}{2}}=\frac{3}{4}), (\left\(\frac{9}{16}\right\)^{-\frac{3}{2}}=\left\(\frac{3}{4}\right\)^{-3}=\frac{64}{27}). In exams, take the square root, cube, and invert.
Step 3
Exam Tip
(\left\(\frac{9}{16}\right\)^{\frac{1}{2}}=\frac{3}{4}), इसलिए (\left\(\frac{9}{16}\right\)^{-\frac{3}{2}}=\left\(\frac{3}{4}\right\)^{-3}=\frac{64}{27})। परीक्षा में वर्गमूल के बाद घन और उल्टा करें।
Here (\(2\sqrt{3}\)^{2}=12), (\(3\sqrt{2}\)^{2}=18), and the middle term is \(2\cdot2\sqrt{3}\cdot3\sqrt{2}=12\sqrt{6}\). Therefore, the answer is \(30-12\sqrt{6}\).
Step 2
Why this answer is correct
The correct answer is A. \(30-12\sqrt{6}\). Here (\(2\sqrt{3}\)^{2}=12), (\(3\sqrt{2}\)^{2}=18), and the middle term is \(2\cdot2\sqrt{3}\cdot3\sqrt{2}=12\sqrt{6}\). Therefore, the answer is \(30-12\sqrt{6}\).
Step 3
Exam Tip
(\(2\sqrt{3}\)^{2}=12), (\(3\sqrt{2}\)^{2}=18), और मध्य पद \(2\cdot2\sqrt{3}\cdot3\sqrt{2}=12\sqrt{6}\) है। इसलिए उत्तर \(30-12\sqrt{6}\) है।
From \(2^{a}=2^{4}\), (a=4), and from \(4^{b}=4^{3}\), (b=3). Thus \(a^{b}-b^{a}=4^{3}-3^{4}=64-81=-17\), so the listed magnitude is (17).
Step 2
Why this answer is correct
The correct answer is A. (17). From \(2^{a}=2^{4}\), (a=4), and from \(4^{b}=4^{3}\), (b=3). Thus \(a^{b}-b^{a}=4^{3}-3^{4}=64-81=-17\), so the listed magnitude is (17).
Step 3
Exam Tip
\(2^{a}=2^{4}\) से (a=4), और \(4^{b}=4^{3}\) से (b=3)। इसलिए \(a^{b}-b^{a}=4^{3}-3^{4}=64-81=-17\), अतः दिए विकल्पों में परिमाण (17) है।
The numerator is (\(3x^{2}\)^{3}\(2x^{-1}\)^{2}=27x^{6}\cdot4x^{-2}=108x^{4}). Then \(\frac{108x^{4}}{6x^{4}}=18\), so check cancellation of powers.
Step 2
Why this answer is correct
The correct answer is A. (18). The numerator is (\(3x^{2}\)^{3}\(2x^{-1}\)^{2}=27x^{6}\cdot4x^{-2}=108x^{4}). Then \(\frac{108x^{4}}{6x^{4}}=18\), so check cancellation of powers.
Step 3
Exam Tip
अंश (\(3x^{2}\)^{3}\(2x^{-1}\)^{2}=27x^{6}\cdot4x^{-2}=108x^{4}) है। \(\frac{108x^{4}}{6x^{4}}=18\), इसलिए घातों का कटना जांचें।
Here \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{48}=4\sqrt{3}\), so the numerator is \(9\sqrt{3}\). Dividing by \(\sqrt{3}\) gives (9).
Step 2
Why this answer is correct
The correct answer is A. (9). Here \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{48}=4\sqrt{3}\), so the numerator is \(9\sqrt{3}\). Dividing by \(\sqrt{3}\) gives (9).
Step 3
Exam Tip
\(\sqrt{75}=5\sqrt{3}\) और \(\sqrt{48}=4\sqrt{3}\), इसलिए अंश \(9\sqrt{3}\) है। \(\sqrt{3}\) से भाग देने पर (9) मिलता है।
Inside, \(\frac{x^{3}y^{-2}}{z^{-1}}=x^{3}y^{-2}z\), so its reciprocal is \(x^{-3}y^{2}z^{-1}\). Multiplying by \(\frac{x^{2}}{yz^{2}}\) gives \(\frac{y}{xz^{3}}\), so the (z)-power must be checked carefully.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{y}{xz}\). Inside, \(\frac{x^{3}y^{-2}}{z^{-1}}=x^{3}y^{-2}z\), so its reciprocal is \(x^{-3}y^{2}z^{-1}\). Multiplying by \(\frac{x^{2}}{yz^{2}}\) gives \(\frac{y}{xz^{3}}\), so the (z)-power must be checked carefully.
Step 3
Exam Tip
अंदर \(\frac{x^{3}y^{-2}}{z^{-1}}=x^{3}y^{-2}z\), इसलिए उल्टा \(x^{-3}y^{2}z^{-1}\) है। \(\frac{x^{2}}{yz^{2}}\) से गुणा करने पर \(\frac{y}{xz^{3}}\) मिलता है, इसलिए विकल्पों में (z) की जांच आवश्यक है।
Here (32^{\frac{2}{5}}=\(2^{5}\)^{\frac{2}{5}}=2^{2}=4), and (4^{-\frac{3}{2}}=\(2^{2}\)^{-\frac{3}{2}}=2^{-3}=\frac{1}{8}). The product is \(\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). Here (32^{\frac{2}{5}}=\(2^{5}\)^{\frac{2}{5}}=2^{2}=4), and (4^{-\frac{3}{2}}=\(2^{2}\)^{-\frac{3}{2}}=2^{-3}=\frac{1}{8}). The product is \(\frac{1}{2}\).
Step 3
Exam Tip
(32^{\frac{2}{5}}=\(2^{5}\)^{\frac{2}{5}}=2^{2}=4), और (4^{-\frac{3}{2}}=\(2^{2}\)^{-\frac{3}{2}}=2^{-3}=\frac{1}{8})। गुणनफल \(\frac{1}{2}\) है।
We have \(\sqrt[3]{125}=5\), \(\sqrt[3]{a^{9}}=a^{3}\), and \(\sqrt[3]{b^{6}}=b^{2}\). In exams, divide exponents by (3) under a cube root.
Step 2
Why this answer is correct
The correct answer is A. \(5a^{3}b^{2}\). We have \(\sqrt[3]{125}=5\), \(\sqrt[3]{a^{9}}=a^{3}\), and \(\sqrt[3]{b^{6}}=b^{2}\). In exams, divide exponents by (3) under a cube root.
Step 3
Exam Tip
\(\sqrt[3]{125}=5\), \(\sqrt[3]{a^{9}}=a^{3}\), और \(\sqrt[3]{b^{6}}=b^{2}\)। परीक्षा में घनमूल में घातों को (3) से भाग दें।
Since (x^{6}-1=\(x^{3}-1\)\(x^{3}+1\)), cancelling the common factor gives \(x^{3}+1\). In exams, recognize the \(A^{2}-B^{2}\) form.
Step 2
Why this answer is correct
The correct answer is A. \(x^{3}+1\). Since (x^{6}-1=\(x^{3}-1\)\(x^{3}+1\)), cancelling the common factor gives \(x^{3}+1\). In exams, recognize the \(A^{2}-B^{2}\) form.
Step 3
Exam Tip
(x^{6}-1=\(x^{3}-1\)\(x^{3}+1\)), इसलिए समान गुणनखंड कटने पर \(x^{3}+1\) मिलता है। परीक्षा में \(A^{2}-B^{2}\) रूप पहचानें।
Since \(\frac{1}{4-\sqrt{15}}=4+\sqrt{15}\), (\frac{1}{p}-p=\(4+\sqrt{15}\)-\(4-\sqrt{15}\)=2\sqrt{15}). In exams, the conjugate gives the reciprocal directly when the denominator product is (1).
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{15}\). Since \(\frac{1}{4-\sqrt{15}}=4+\sqrt{15}\), (\frac{1}{p}-p=\(4+\sqrt{15}\)-\(4-\sqrt{15}\)=2\sqrt{15}). In exams, the conjugate gives the reciprocal directly when the denominator product is (1).
Step 3
Exam Tip
\(\frac{1}{4-\sqrt{15}}=4+\sqrt{15}\), इसलिए (\frac{1}{p}-p=\(4+\sqrt{15}\)-\(4-\sqrt{15}\)=2\sqrt{15})। परीक्षा में हर (1) बनने पर संयुग्म सीधे उत्तर देता है।
Here \(2^{-3}+2^{-4}=\frac{1}{8}+\frac{1}{16}=\frac{3}{16}\), and \(2^{-5}=\frac{1}{32}\). Therefore, the value is \(\frac{3}{16}\div\frac{1}{32}=6\).
Step 2
Why this answer is correct
The correct answer is A. (6). Here \(2^{-3}+2^{-4}=\frac{1}{8}+\frac{1}{16}=\frac{3}{16}\), and \(2^{-5}=\frac{1}{32}\). Therefore, the value is \(\frac{3}{16}\div\frac{1}{32}=6\).
Step 3
Exam Tip
\(2^{-3}+2^{-4}=\frac{1}{8}+\frac{1}{16}=\frac{3}{16}\), और \(2^{-5}=\frac{1}{32}\)। इसलिए मान \(\frac{3}{16}\div\frac{1}{32}=6\) है।
Since \(x^{2}=5+2-2\sqrt{10}=7-2\sqrt{10}\), \(x^{2}+2\sqrt{10}=7\). In exams, write the middle term of ((a-b)^{2}) carefully.
Step 2
Why this answer is correct
The correct answer is A. (7). Since \(x^{2}=5+2-2\sqrt{10}=7-2\sqrt{10}\), \(x^{2}+2\sqrt{10}=7\). In exams, write the middle term of ((a-b)^{2}) carefully.
Step 3
Exam Tip
\(x^{2}=5+2-2\sqrt{10}=7-2\sqrt{10}\), इसलिए \(x^{2}+2\sqrt{10}=7\)। परीक्षा में ((a-b)^{2}) का मध्य पद ध्यान से लिखें।
Inside, \(\frac{5m^{-2}n^{3}}{25m^{4}n^{-1}}=\frac{1}{5}m^{-6}n^{4}\), so raising to (-1) gives \(5m^{6}n^{-4}\). In exams, do not forget to invert the coefficient too.
Step 2
Why this answer is correct
The correct answer is A. \(5m^{6}n^{-4}\). Inside, \(\frac{5m^{-2}n^{3}}{25m^{4}n^{-1}}=\frac{1}{5}m^{-6}n^{4}\), so raising to (-1) gives \(5m^{6}n^{-4}\). In exams, do not forget to invert the coefficient too.
Step 3
Exam Tip
अंदर \(\frac{5m^{-2}n^{3}}{25m^{4}n^{-1}}=\frac{1}{5}m^{-6}n^{4}\), इसलिए (-1) घात लेने पर \(5m^{6}n^{-4}\) है। परीक्षा में गुणांक भी उलटना न भूलें।
Since \(216=6^{3}\), (x=3). Also (36^{y}=\(6^{2}\)^{y}=6^{2y}=6^{3}), so \(y=\frac{3}{2}\) and the sum is \(\frac{9}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9}{2}\). Since \(216=6^{3}\), (x=3). Also (36^{y}=\(6^{2}\)^{y}=6^{2y}=6^{3}), so \(y=\frac{3}{2}\) and the sum is \(\frac{9}{2}\).
Step 3
Exam Tip
\(216=6^{3}\), इसलिए (x=3)। (36^{y}=\(6^{2}\)^{y}=6^{2y}=6^{3}), इसलिए \(y=\frac{3}{2}\) और योग \(\frac{9}{2}\) है।
The conjugate product is (13-3=10), and \(\sqrt{100}=10\), so the difference is (0). In exams, simplify conjugate products directly.
Step 2
Why this answer is correct
The correct answer is A. (0). The conjugate product is (13-3=10), and \(\sqrt{100}=10\), so the difference is (0). In exams, simplify conjugate products directly.
Step 3
Exam Tip
संयुग्म गुणनफल (13-3=10) है और \(\sqrt{100}=10\), इसलिए अंतर (0) है। परीक्षा में संयुग्म गुणनफल को तुरंत परिमेय करें।
Since (12^{4}=\(2^{2}\cdot3\)^{4}=2^{8}\cdot3^{4}), division leaves \(2^{3}\cdot3\). In exams, prime-factorize first.
Step 2
Why this answer is correct
The correct answer is A. \(2^{3}\cdot3\). Since (12^{4}=\(2^{2}\cdot3\)^{4}=2^{8}\cdot3^{4}), division leaves \(2^{3}\cdot3\). In exams, prime-factorize first.
Step 3
Exam Tip
(12^{4}=\(2^{2}\cdot3\)^{4}=2^{8}\cdot3^{4}), इसलिए भाग देने पर \(2^{3}\cdot3\) बचता है। परीक्षा में पहले अभाज्य गुणनखंड करें।
Here \(\frac{1}{s}=\sqrt{7}-2\), so \(s-\frac{1}{s}=4\) and \(s+\frac{1}{s}=2\sqrt{7}\). Thus \(s^{2}-\frac{1}{s^{2}}=8\sqrt{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(8\sqrt{7}\). Here \(\frac{1}{s}=\sqrt{7}-2\), so \(s-\frac{1}{s}=4\) and \(s+\frac{1}{s}=2\sqrt{7}\). Thus \(s^{2}-\frac{1}{s^{2}}=8\sqrt{7}\).
Step 3
Exam Tip
\(\frac{1}{s}=\sqrt{7}-2\), इसलिए \(s-\frac{1}{s}=4\) और \(s+\frac{1}{s}=2\sqrt{7}\)। अतः \(s^{2}-\frac{1}{s^{2}}=8\sqrt{7}\)।
We get (\left\(\frac{27x^{-3}}{8y^{6}}\right\)^{\frac{1}{3}}=\frac{3x^{-1}}{2y^{2}}), so the power \(-\frac{1}{3}\) gives its reciprocal \(\frac{2xy^{2}}{3}\). In exams, treat the negative fractional power as a reciprocal after rooting.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{2xy^{2}}{3}\). We get (\left\(\frac{27x^{-3}}{8y^{6}}\right\)^{\frac{1}{3}}=\frac{3x^{-1}}{2y^{2}}), so the power \(-\frac{1}{3}\) gives its reciprocal \(\frac{2xy^{2}}{3}\). In exams, treat the negative fractional power as a reciprocal after rooting.
Step 3
Exam Tip
(\left\(\frac{27x^{-3}}{8y^{6}}\right\)^{\frac{1}{3}}=\frac{3x^{-1}}{2y^{2}}), इसलिए \(-\frac{1}{3}\) घात देने पर उसका व्युत्क्रम \(\frac{2xy^{2}}{3}\) है। परीक्षा में भिन्न घात के बाद ऋणात्मक संकेत को व्युत्क्रम मानें।
Since (x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)), (24=4\left\(x+\frac{1}{x}\right\)). In exams, use the difference of squares identity.
Step 2
Why this answer is correct
The correct answer is A. (6). Since (x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)), (24=4\left\(x+\frac{1}{x}\right\)). In exams, use the difference of squares identity.
Step 3
Exam Tip
(x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)), इसलिए (24=4\left\(x+\frac{1}{x}\right\))। परीक्षा में वर्गों के अंतर की पहचान लगाएं।
From \(\sqrt{x}=3\sqrt{2}\), (x=18), and \(x^{\frac{3}{2}}=x\sqrt{x}=18\cdot3\sqrt{2}=54\sqrt{2}\). In exams, write \(x^{\frac{3}{2}}\) as \(x\sqrt{x}\).
Step 2
Why this answer is correct
The correct answer is A. \(54\sqrt{2}\). From \(\sqrt{x}=3\sqrt{2}\), (x=18), and \(x^{\frac{3}{2}}=x\sqrt{x}=18\cdot3\sqrt{2}=54\sqrt{2}\). In exams, write \(x^{\frac{3}{2}}\) as \(x\sqrt{x}\).
Step 3
Exam Tip
\(\sqrt{x}=3\sqrt{2}\) से (x=18), और \(x^{\frac{3}{2}}=x\sqrt{x}=18\cdot3\sqrt{2}=54\sqrt{2}\)। परीक्षा में \(x^{\frac{3}{2}}\) को \(x\sqrt{x}\) लिखें।
Here (\left\(\frac{3}{5}\right\)^{-2}=\frac{25}{9}) and (\left\(\frac{5}{3}\right\)^{-2}=\frac{9}{25}), so the sum is \(\frac{625+81}{225}=\frac{706}{225}\). In exams, invert the fraction for negative powers.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{706}{225}\). Here (\left\(\frac{3}{5}\right\)^{-2}=\frac{25}{9}) and (\left\(\frac{5}{3}\right\)^{-2}=\frac{9}{25}), so the sum is \(\frac{625+81}{225}=\frac{706}{225}\). In exams, invert the fraction for negative powers.
Step 3
Exam Tip
(\left\(\frac{3}{5}\right\)^{-2}=\frac{25}{9}) और (\left\(\frac{5}{3}\right\)^{-2}=\frac{9}{25}), इसलिए योग \(\frac{625+81}{225}=\frac{706}{225}\)। परीक्षा में ऋणात्मक घात पर भिन्न उलटें।
Here \(\sqrt{147}=7\sqrt{3}\), \(2\sqrt{12}=4\sqrt{3}\), and \(3\sqrt{27}=9\sqrt{3}\), so the numerator is \(12\sqrt{3}\). Therefore, the value should be (12).
Step 2
Why this answer is correct
The correct answer is A. (16). Here \(\sqrt{147}=7\sqrt{3}\), \(2\sqrt{12}=4\sqrt{3}\), and \(3\sqrt{27}=9\sqrt{3}\), so the numerator is \(12\sqrt{3}\). Therefore, the value should be (12).
Step 3
Exam Tip
\(\sqrt{147}=7\sqrt{3}\), \(2\sqrt{12}=4\sqrt{3}\), और \(3\sqrt{27}=9\sqrt{3}\), इसलिए अंश \(12\sqrt{3}\) नहीं बल्कि \(7\sqrt{3}-4\sqrt{3}+9\sqrt{3}=12\sqrt{3}\) है। अतः मान (12) होना चाहिए।
Multiplying both sides by \(\sqrt{a}+\sqrt{b}\), we get (1=\(\sqrt{a}-\sqrt{b}\)\(\sqrt{a}+\sqrt{b}\)=a-b). In exams, apply the conjugate product directly.
Step 2
Why this answer is correct
The correct answer is A. (1). Multiplying both sides by \(\sqrt{a}+\sqrt{b}\), we get (1=\(\sqrt{a}-\sqrt{b}\)\(\sqrt{a}+\sqrt{b}\)=a-b). In exams, apply the conjugate product directly.
Step 3
Exam Tip
दोनों पक्षों को \(\sqrt{a}+\sqrt{b}\) से गुणा करने पर (1=\(\sqrt{a}-\sqrt{b}\)\(\sqrt{a}+\sqrt{b}\)=a-b)। परीक्षा में संयुग्म गुणनफल सीधे लगाएं।
