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With rational coefficients, \(a+\sqrt{b}\) is accompanied by \(a-\sqrt{b}\). In exams identify conjugate zeroes quickly.
Step 2
Why this answer is correct
The correct answer is A. \(2-\sqrt{7}\). With rational coefficients, \(a+\sqrt{b}\) is accompanied by \(a-\sqrt{b}\). In exams identify conjugate zeroes quickly.
Step 3
Exam Tip
परिमेय गुणांकों में \(a+\sqrt{b}\) के साथ \(a-\sqrt{b}\) भी शून्यक आता है। परीक्षा में संयुग्मी शून्यकों को तुरंत पहचानें।
The denominator contains (7), so the decimal will not terminate and being rational it will recur. In exams decide after checking the denominator in lowest form.
Step 2
Why this answer is correct
The correct answer is A. अनवसानी आवर्ती / Non-terminating recurring. The denominator contains (7), so the decimal will not terminate and being rational it will recur. In exams decide after checking the denominator in lowest form.
Step 3
Exam Tip
हर में (7) है, इसलिए दशमलव समाप्त नहीं होगा और परिमेय होने से आवर्ती होगा। परीक्षा में हर को सरलतम रूप में देखकर निर्णय लें।
In \(\frac{63}{2^5\cdot5^2\cdot7}\), after cancelling (63) and (7), only (2) and (5) remain in the denominator. In exams reduce the fraction first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{63}{2^5\cdot5^2\cdot7}\). In \(\frac{63}{2^5\cdot5^2\cdot7}\), after cancelling (63) and (7), only (2) and (5) remain in the denominator. In exams reduce the fraction first.
Step 3
Exam Tip
\(\frac{63}{2^5\cdot5^2\cdot7}\) में (63) और (7) कटने के बाद हर में केवल (2) और (5) बचते हैं। परीक्षा में पहले भिन्न को सरलतम रूप में लाएं।
Multiplying the denominator by \(\sqrt{5}+2\) makes it (5-4=1). In exams choose the conjugate of the denominator.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{5}+2\). Multiplying the denominator by \(\sqrt{5}+2\) makes it (5-4=1). In exams choose the conjugate of the denominator.
Step 3
Exam Tip
हर को \(\sqrt{5}+2\) से गुणा करने पर हर (5-4=1) बनता है। परीक्षा में हर का संयुग्मी चुनें।
A. (m) पूर्ण वर्ग नहीं है/(m) is not a perfect square
Step 1
Concept
The square root of a perfect square is an integer, so for an irrational square root (m) is not a perfect square. In exams identifying perfect squares is important.
Step 2
Why this answer is correct
The correct answer is A. (m) पूर्ण वर्ग नहीं है / (m) is not a perfect square. The square root of a perfect square is an integer, so for an irrational square root (m) is not a perfect square. In exams identifying perfect squares is important.
Step 3
Exam Tip
पूर्ण वर्ग का वर्गमूल पूर्णांक होता है, इसलिए अपरिमेय वर्गमूल के लिए (m) पूर्ण वर्ग नहीं होगा। परीक्षा में पूर्ण वर्ग पहचानना जरूरी है।
Rationalizing \(\frac{1}{2-\sqrt{3}}\) with \(2+\sqrt{3}\) gives \(2+\sqrt{3}\). In exams multiply by the conjugate of the denominator.
Step 2
Why this answer is correct
The correct answer is A. \(2+\sqrt{3}\). Rationalizing \(\frac{1}{2-\sqrt{3}}\) with \(2+\sqrt{3}\) gives \(2+\sqrt{3}\). In exams multiply by the conjugate of the denominator.
Step 3
Exam Tip
\(\frac{1}{2-\sqrt{3}}\) को \(2+\sqrt{3}\) से परिमेयकृत करने पर \(2+\sqrt{3}\) मिलता है। परीक्षा में हर का संयुग्मी लगाएं।
(4) is rational and \(\sqrt{13}\) is irrational, so the sum is irrational. In exams identify square roots of perfect squares first.
Step 2
Why this answer is correct
The correct answer is A. \(4+\sqrt{13}\). (4) is rational and \(\sqrt{13}\) is irrational, so the sum is irrational. In exams identify square roots of perfect squares first.
Step 3
Exam Tip
(4) परिमेय है और \(\sqrt{13}\) अपरिमेय है, इसलिए योग अपरिमेय है। परीक्षा में पूर्ण वर्ग के वर्गमूल को पहले पहचानें।
A. \(2+\sqrt{5}\) और \(2-\sqrt{5}\)/\(2+\sqrt{5}\) and \(2-\sqrt{5}\)
Step 1
Concept
The sum is (\(2+\sqrt{5}\)+\(2-\sqrt{5}\)=4), which is rational. In exams remember conjugate pairs as counterexamples.
Step 2
Why this answer is correct
The correct answer is A. \(2+\sqrt{5}\) और \(2-\sqrt{5}\) / \(2+\sqrt{5}\) and \(2-\sqrt{5}\). The sum is (\(2+\sqrt{5}\)+\(2-\sqrt{5}\)=4), which is rational. In exams remember conjugate pairs as counterexamples.
Step 3
Exam Tip
योग (\(2+\sqrt{5}\)+\(2-\sqrt{5}\)=4) है, जो परिमेय है। परीक्षा में संयुग्मी जोड़ों को प्रतिउदाहरण के रूप में याद रखें।
A. (p) और (q) दोनों (3) से विभाज्य हो जाते हैं/Both (p) and (q) become divisible by (3)
Step 1
Concept
From \(\sqrt{3}=\frac{p}{q}\), we get \(p^2=3q^2\), so both (p) and (q) become divisible by (3). In exams use the coprime condition at the end.
Step 2
Why this answer is correct
The correct answer is A. (p) और (q) दोनों (3) से विभाज्य हो जाते हैं / Both (p) and (q) become divisible by (3). From \(\sqrt{3}=\frac{p}{q}\), we get \(p^2=3q^2\), so both (p) and (q) become divisible by (3). In exams use the coprime condition at the end.
Step 3
Exam Tip
\(\sqrt{3}=\frac{p}{q}\) से \(p^2=3q^2\) मिलता है, इसलिए (p) और (q) दोनों (3) से विभाज्य हो जाते हैं। परीक्षा में सहअभाज्य शर्त को अंत में उपयोग करें।
The sum is (2) and the product is (1-6=-5), so the polynomial is \(x^2-2x-5\). In exams use \(a^2-b^2\) for the product.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-2x-5\). The sum is (2) and the product is (1-6=-5), so the polynomial is \(x^2-2x-5\). In exams use \(a^2-b^2\) for the product.
Step 3
Exam Tip
योग (2) और गुणनफल (1-6=-5) है, इसलिए बहुपद \(x^2-2x-5\) है। परीक्षा में गुणनफल में \(a^2-b^2\) लगाएं।
A. \(2+\sqrt{5}\) और \(2-\sqrt{5}\)/\(2+\sqrt{5}\) and \(2-\sqrt{5}\)
Step 1
Concept
Using the quadratic formula, \(x=\frac{4\pm\sqrt{16+4}}{2}=2\pm\sqrt{5}\). In exams simplify the discriminant.
Step 2
Why this answer is correct
The correct answer is A. \(2+\sqrt{5}\) और \(2-\sqrt{5}\) / \(2+\sqrt{5}\) and \(2-\sqrt{5}\). Using the quadratic formula, \(x=\frac{4\pm\sqrt{16+4}}{2}=2\pm\sqrt{5}\). In exams simplify the discriminant.
Step 3
Exam Tip
द्विघात सूत्र से \(x=\frac{4\pm\sqrt{16+4}}{2}=2\pm\sqrt{5}\) मिलता है। परीक्षा में विविक्तकर को सरल करें।
\(\sqrt{-9}\) is not a real number, while the others are real. In exams do not take the square root of a negative number in the real number system.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{-9}\). \(\sqrt{-9}\) is not a real number, while the others are real. In exams do not take the square root of a negative number in the real number system.
Step 3
Exam Tip
\(\sqrt{-9}\) वास्तविक संख्या नहीं है, जबकि बाकी सभी वास्तविक हैं। परीक्षा में ऋणात्मक संख्या का वर्गमूल वास्तविक संख्या पद्धति में नहीं लेते।
\(\frac{1}{3+\sqrt{10}}=\sqrt{10}-3\), so the sum is \(2\sqrt{10}\). In exams rationalize the reciprocal first.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{10}\). \(\frac{1}{3+\sqrt{10}}=\sqrt{10}-3\), so the sum is \(2\sqrt{10}\). In exams rationalize the reciprocal first.
Step 3
Exam Tip
\(\frac{1}{3+\sqrt{10}}=\sqrt{10}-3\), इसलिए योग \(2\sqrt{10}\) है। परीक्षा में पहले व्युत्क्रम को परिमेयकृत करें।
After simplification, (7) remains in the denominator, so the decimal is non-terminating recurring. In exams do not decide only from the original denominator.
Step 2
Why this answer is correct
The correct answer is A. अनवसानी आवर्ती / Non-terminating recurring. After simplification, (7) remains in the denominator, so the decimal is non-terminating recurring. In exams do not decide only from the original denominator.
Step 3
Exam Tip
सरलीकरण के बाद हर में (7) बचता है, इसलिए दशमलव अनवसानी आवर्ती होगा। परीक्षा में केवल मूल हर देखकर निर्णय न लें।
\(\sqrt{20}+\sqrt{45}=2\sqrt{5}+3\sqrt{5}=5\sqrt{5}\), which is irrational. In exams do not treat addition like multiplication.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{20}+\sqrt{45}\). \(\sqrt{20}+\sqrt{45}=2\sqrt{5}+3\sqrt{5}=5\sqrt{5}\), which is irrational. In exams do not treat addition like multiplication.
Step 3
Exam Tip
\(\sqrt{20}+\sqrt{45}=2\sqrt{5}+3\sqrt{5}=5\sqrt{5}\), जो अपरिमेय है। परीक्षा में योग को गुणन जैसा न मानें।
The discriminant is (100-92=8), and \(\sqrt{8}\) is irrational, so the zeroes are real irrational. In exams check the square root of the discriminant.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक अपरिमेय / Real irrational. The discriminant is (100-92=8), and \(\sqrt{8}\) is irrational, so the zeroes are real irrational. In exams check the square root of the discriminant.
Step 3
Exam Tip
विविक्तकर (100-92=8) है और \(\sqrt{8}\) अपरिमेय है, इसलिए शून्यक वास्तविक अपरिमेय हैं। परीक्षा में विविक्तकर का वर्गमूल देखें।
\(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{32}=4\sqrt{2}\), so the value is \(2\sqrt{2}\). In exams handle signs carefully.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{2}\). \(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{32}=4\sqrt{2}\), so the value is \(2\sqrt{2}\). In exams handle signs carefully.
Step 3
Exam Tip
\(\sqrt{50}=5\sqrt{2}\) और \(\sqrt{32}=4\sqrt{2}\), इसलिए मान \(2\sqrt{2}\) है। परीक्षा में चिन्हों को सावधानी से रखें।
A. \(4+\sqrt{5}\) और \(4-\sqrt{5}\)/\(4+\sqrt{5}\) and \(4-\sqrt{5}\)
Step 1
Concept
The sum of \(4+\sqrt{5}\) and \(4-\sqrt{5}\) is (8), and the product is (16-5=11). In exams check the sum and product of options.
Step 2
Why this answer is correct
The correct answer is A. \(4+\sqrt{5}\) और \(4-\sqrt{5}\) / \(4+\sqrt{5}\) and \(4-\sqrt{5}\). The sum of \(4+\sqrt{5}\) and \(4-\sqrt{5}\) is (8), and the product is (16-5=11). In exams check the sum and product of options.
Step 3
Exam Tip
\(4+\sqrt{5}\) और \(4-\sqrt{5}\) का योग (8) और गुणनफल (16-5=11) है। परीक्षा में विकल्पों का योग और गुणनफल जांचें।
The principal square root is always non-negative, so \(\sqrt{a^2}=|a|\). In exams do not forget the possibility of negative (a).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{a^2}=|a|\). The principal square root is always non-negative, so \(\sqrt{a^2}=|a|\). In exams do not forget the possibility of negative (a).
Step 3
Exam Tip
मुख्य वर्गमूल हमेशा अऋणात्मक होता है, इसलिए \(\sqrt{a^2}=|a|\) है। परीक्षा में (a) ऋणात्मक होने की संभावना न भूलें।
(x-2-y-2=(x-y)(x+y)=\(2\sqrt{2}\)\(2\sqrt{5}\)=4\sqrt{10}). In exams use identities to avoid long calculation.
Step 2
Why this answer is correct
The correct answer is A. \(4\sqrt{10}\). (x-2-y-2=(x-y)(x+y)=\(2\sqrt{2}\)\(2\sqrt{5}\)=4\sqrt{10}). In exams use identities to avoid long calculation.
Step 3
Exam Tip
(x-2-y-2=(x-y)(x+y)=\(2\sqrt{2}\)\(2\sqrt{5}\)=4\sqrt{10}) है। परीक्षा में पहचान का प्रयोग करके लंबी गणना बचाएं।
\(\sqrt{17}\) is irrational, so its decimal expansion is non-terminating non-recurring. In exams distinguish irrational decimals from recurring decimals.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{17}\). \(\sqrt{17}\) is irrational, so its decimal expansion is non-terminating non-recurring. In exams distinguish irrational decimals from recurring decimals.
Step 3
Exam Tip
\(\sqrt{17}\) अपरिमेय है, इसलिए इसका दशमलव अनवसानी अनावर्ती होगा। परीक्षा में अपरिमेय और आवर्ती दशमलव में अंतर रखें।
The like (x) terms cancel and the value left is \(2\sqrt{2}\). In exams do not be confused by the type of number during algebraic simplification.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{2}\). The like (x) terms cancel and the value left is \(2\sqrt{2}\). In exams do not be confused by the type of number during algebraic simplification.
Step 3
Exam Tip
समान (x) पद कट जाते हैं और मान \(2\sqrt{2}\) बचता है। परीक्षा में बीजीय सरलीकरण में संख्या के प्रकार से भ्रमित न हों।
A. \(\sqrt{2}\) और \(3\sqrt{2}\)/\(\sqrt{2}\) and \(3\sqrt{2}\)
Step 1
Concept
\(\sqrt{2}\cdot3\sqrt{2}=6\), which is rational. In exams remember counterexamples for products of irrational numbers.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\) और \(3\sqrt{2}\) / \(\sqrt{2}\) and \(3\sqrt{2}\). \(\sqrt{2}\cdot3\sqrt{2}=6\), which is rational. In exams remember counterexamples for products of irrational numbers.
Step 3
Exam Tip
\(\sqrt{2}\cdot3\sqrt{2}=6\), जो परिमेय है। परीक्षा में अपरिमेय संख्याओं के गुणनफल के लिए प्रतिउदाहरण याद रखें।
The companion zero is \(2-\sqrt{3}\), so the factor is (x-\(2-\sqrt{3}\)). In exams remember the relation between a zero and factor as \(x-\alpha\).
Step 2
Why this answer is correct
The correct answer is A. (x-\(2-\sqrt{3}\)). The companion zero is \(2-\sqrt{3}\), so the factor is (x-\(2-\sqrt{3}\)). In exams remember the relation between a zero and factor as \(x-\alpha\).
Step 3
Exam Tip
साथी शून्यक \(2-\sqrt{3}\) होगा, इसलिए गुणनखंड (x-\(2-\sqrt{3}\)) है। परीक्षा में शून्यक और गुणनखंड का संबंध \(x-\alpha\) याद रखें।
\(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the square is (\(5\sqrt{3}\)2=75); none of the expanded radical options except the simplified value idea fits. In exams simplify before expanding.
Step 2
Why this answer is correct
The correct answer is A. \(75+36\sqrt{3}\). \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the square is (\(5\sqrt{3}\)2=75); none of the expanded radical options except the simplified value idea fits. In exams simplify before expanding.
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{27}=3\sqrt{3}\), इसलिए वर्ग (\(5\sqrt{3}\)2=75) होना चाहिए, पर विकल्पों में विस्तार विधि से सही मान (75) अकेला नहीं है। परीक्षा में पहले सरलीकरण करें।
The sum is (12) and the product is (36-11=25), so the polynomial is \(x^2-12x+25\). In exams write the standard form correctly.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-12x+25\). The sum is (12) and the product is (36-11=25), so the polynomial is \(x^2-12x+25\). In exams write the standard form correctly.
Step 3
Exam Tip
योग (12) और गुणनफल (36-11=25) है, इसलिए बहुपद \(x^2-12x+25\) है। परीक्षा में मानक रूप ठीक से लिखें।
The conjugate of the denominator is \(\sqrt{7}-\sqrt{6}\), and the denominator becomes (7-6=1). In exams the answer simplifies when the difference is (1).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{7}-\sqrt{6}\). The conjugate of the denominator is \(\sqrt{7}-\sqrt{6}\), and the denominator becomes (7-6=1). In exams the answer simplifies when the difference is (1).
Step 3
Exam Tip
हर का संयुग्मी \(\sqrt{7}-\sqrt{6}\) है और हर (7-6=1) बनता है। परीक्षा में अंतर (1) होने पर उत्तर सरल हो जाता है।
A. \(\sqrt{8}\) और \(-\sqrt{8}\)/\(\sqrt{8}\) and \(-\sqrt{8}\)
Step 1
Concept
(\sqrt{8}+\(-\sqrt{8}\)=0), which is rational. In exams one counterexample is enough to disprove a universal statement.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{8}\) और \(-\sqrt{8}\) / \(\sqrt{8}\) and \(-\sqrt{8}\). (\sqrt{8}+\(-\sqrt{8}\)=0), which is rational. In exams one counterexample is enough to disprove a universal statement.
Step 3
Exam Tip
(\sqrt{8}+\(-\sqrt{8}\)=0), जो परिमेय है। परीक्षा में गलत सार्वत्रिक कथन तोड़ने के लिए एक प्रतिउदाहरण काफी है।
If \(q\ne0\), then \(q\sqrt{5}\) is irrational and the sum cannot be rational. In exams check the possibility of a zero coefficient.
Step 2
Why this answer is correct
The correct answer is A. (q=0). If \(q\ne0\), then \(q\sqrt{5}\) is irrational and the sum cannot be rational. In exams check the possibility of a zero coefficient.
Step 3
Exam Tip
यदि \(q\ne0\), तो \(q\sqrt{5}\) अपरिमेय होगा और योग परिमेय नहीं हो सकता। परीक्षा में शून्य गुणांक की संभावना देखें।
Multiplying by the conjugate of the denominator gives denominator (1) and numerator (\(\sqrt{3}+\sqrt{2}\)2=5+2\sqrt{6}). In exams apply the conjugate in one step.
Step 2
Why this answer is correct
The correct answer is A. \(5+2\sqrt{6}\). Multiplying by the conjugate of the denominator gives denominator (1) and numerator (\(\sqrt{3}+\sqrt{2}\)2=5+2\sqrt{6}). In exams apply the conjugate in one step.
Step 3
Exam Tip
हर के संयुग्मी से गुणा करने पर हर (1) और अंश (\(\sqrt{3}+\sqrt{2}\)2=5+2\sqrt{6}) बनता है। परीक्षा में एक ही चरण में संयुग्मी लगाएं।
From the product \(k^2-7=9\), we get \(k^2=16\), and (k=4) fits the given form. In exams use the product to find the unknown.
Step 2
Why this answer is correct
The correct answer is A. (4). From the product \(k^2-7=9\), we get \(k^2=16\), and (k=4) fits the given form. In exams use the product to find the unknown.
Step 3
Exam Tip
गुणनफल \(k^2-7=9\) से \(k^2=16\) मिलता है और दिए रूप में (k=4) उपयुक्त है। परीक्षा में गुणनफल से अज्ञात निकालें।
A. \(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\) हमेशा/\(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\) always
Step 1
Concept
\(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\) is generally false, for example \(\sqrt{9+16}\ne3+4\). In exams do not split addition inside a radical.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\) हमेशा / \(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\) always. \(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\) is generally false, for example \(\sqrt{9+16}\ne3+4\). In exams do not split addition inside a radical.
Step 3
Exam Tip
\(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\) सामान्यतः गलत है, जैसे \(\sqrt{9+16}\ne3+4\)। परीक्षा में मूल के अंदर योग को अलग-अलग न तोड़ें।
Since \(2<\frac{5}{2}<3\), \(\sqrt{2}<\sqrt{\frac{5}{2}}<\sqrt{3}\), and it is irrational. In exams compare by squaring.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{\frac{5}{2}}\). Since \(2<\frac{5}{2}<3\), \(\sqrt{2}<\sqrt{\frac{5}{2}}<\sqrt{3}\), and it is irrational. In exams compare by squaring.
Step 3
Exam Tip
क्योंकि \(2<\frac{5}{2}<3\), इसलिए \(\sqrt{2}<\sqrt{\frac{5}{2}}<\sqrt{3}\) और यह अपरिमेय है। परीक्षा में वर्ग करके तुलना करें।
The companion zero is \(5-2\sqrt{6}\), with sum (10) and product (25-24=1). In exams form the polynomial using the conjugate.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-10x+1\). The companion zero is \(5-2\sqrt{6}\), with sum (10) and product (25-24=1). In exams form the polynomial using the conjugate.
Step 3
Exam Tip
साथी शून्यक \(5-2\sqrt{6}\) होगा, योग (10) और गुणनफल (25-24=1) है। परीक्षा में संयुग्मी लेकर बहुपद बनाएं।
The product (\(3+\sqrt{2}\)\(3-\sqrt{2}\)=9-2=7), so (k=7). In exams connect the constant term with the product of zeroes.
Step 2
Why this answer is correct
The correct answer is A. (7). The product (\(3+\sqrt{2}\)\(3-\sqrt{2}\)=9-2=7), so (k=7). In exams connect the constant term with the product of zeroes.
Step 3
Exam Tip
गुणनफल (\(3+\sqrt{2}\)\(3-\sqrt{2}\)=9-2=7) है, इसलिए (k=7) होगा। परीक्षा में स्थिर पद को शून्यकों के गुणनफल से जोड़ें।
Multiplying by the conjugate \(\sqrt{11}+3\) makes the denominator (11-9=2), and (2) cancels. In exams choose the conjugate of the denominator correctly.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{11}+3\). Multiplying by the conjugate \(\sqrt{11}+3\) makes the denominator (11-9=2), and (2) cancels. In exams choose the conjugate of the denominator correctly.
Step 3
Exam Tip
हर के संयुग्मी \(\sqrt{11}+3\) से गुणा करने पर हर (11-9=2) बनता है और (2) कट जाता है। परीक्षा में हर का संयुग्मी सही चुनें।
Since (\(\sqrt{13}+\sqrt{12}\)\(\sqrt{13}-\sqrt{12}\)=1), the reciprocal is \(\sqrt{13}-\sqrt{12}\). In exams quickly identify conjugates where (a-b=1).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{13}-\sqrt{12}\). Since (\(\sqrt{13}+\sqrt{12}\)\(\sqrt{13}-\sqrt{12}\)=1), the reciprocal is \(\sqrt{13}-\sqrt{12}\). In exams quickly identify conjugates where (a-b=1).
Step 3
Exam Tip
क्योंकि (\(\sqrt{13}+\sqrt{12}\)\(\sqrt{13}-\sqrt{12}\)=1), इसलिए व्युत्क्रम \(\sqrt{13}-\sqrt{12}\) है। परीक्षा में (a-b=1) वाले संयुग्मी जल्दी पहचानें।
\(154=2\cdot7\cdot11\), so after cancellation only \(5^2\) remains in the denominator. In exams decide from the denominator in lowest form.
Step 2
Why this answer is correct
The correct answer is A. समाप्त दशमलव / Terminating decimal. \(154=2\cdot7\cdot11\), so after cancellation only \(5^2\) remains in the denominator. In exams decide from the denominator in lowest form.
Step 3
Exam Tip
\(154=2\cdot7\cdot11\), इसलिए कटने के बाद हर में केवल \(5^2\) बचता है। परीक्षा में निर्णय हमेशा सरलतम रूप के हर से करें।
