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Here \(9^{-1}=3^{-2}\) and \(27^{-1}=3^{-3}\), so the value is \(3^{4-2-(-3)}=3^5=243\). In exams, convert all terms to the same base.
Step 2
Why this answer is correct
The correct answer is A. (,243,). Here \(9^{-1}=3^{-2}\) and \(27^{-1}=3^{-3}\), so the value is \(3^{4-2-(-3)}=3^5=243\). In exams, convert all terms to the same base.
Step 3
Exam Tip
यहां \(9^{-1}=3^{-2}\) और \(27^{-1}=3^{-3}\), इसलिए मान \(3^{4-2-(-3)}=3^5=243\) है। परीक्षा में सभी पदों को समान आधार में बदलें।
The numerator is (\(a^{-2}b^3\)2=a^{-4}b-6) and the denominator is (\(ab^{-1}\)^{-1}=a^{-1}b), so the answer is \(\dfrac{b^5}{a^3}\). In exams, apply the outside power first.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{b^5}{a^3},\). The numerator is (\(a^{-2}b^3\)2=a^{-4}b-6) and the denominator is (\(ab^{-1}\)^{-1}=a^{-1}b), so the answer is \(\dfrac{b^5}{a^3}\). In exams, apply the outside power first.
Step 3
Exam Tip
ऊपर (\(a^{-2}b^3\)2=a^{-4}b-6) और नीचे (\(ab^{-1}\)^{-1}=a^{-1}b), इसलिए उत्तर \(\dfrac{b^5}{a^3}\) है। परीक्षा में बाहर की घात पहले लगाएं।
Since \(25^{\frac{3}{2}}=125\) and \(125^{\frac{2}{3}}=25\), the value is (5). In exams, understand the root first in fractional powers.
Step 2
Why this answer is correct
The correct answer is A. (,5,). Since \(25^{\frac{3}{2}}=125\) and \(125^{\frac{2}{3}}=25\), the value is (5). In exams, understand the root first in fractional powers.
Step 3
Exam Tip
क्योंकि \(25^{\frac{3}{2}}=125\) और \(125^{\frac{2}{3}}=25\), इसलिए मान (5) है। परीक्षा में fractional powers में पहले root समझें।
\(\sqrt{75}=5\sqrt{3}\), \(\sqrt{12}=2\sqrt{3}\), and \(\sqrt{48}=4\sqrt{3}\), so the answer is \(7\sqrt{3}\). In exams, combine only terms with the same radical part.
Step 2
Why this answer is correct
The correct answer is A. \(,7\sqrt{3},\). \(\sqrt{75}=5\sqrt{3}\), \(\sqrt{12}=2\sqrt{3}\), and \(\sqrt{48}=4\sqrt{3}\), so the answer is \(7\sqrt{3}\). In exams, combine only terms with the same radical part.
Step 3
Exam Tip
\(\sqrt{75}=5\sqrt{3}\), \(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{48}=4\sqrt{3}\), इसलिए उत्तर \(7\sqrt{3}\) है। परीक्षा में समान मूल वाले पद ही जोड़ें।
Multiplying by \(\sqrt{7}-\sqrt{5}\) makes the denominator (7-5=2) and gives \(\sqrt{7}-\sqrt{5}\). In exams, use the conjugate.
Step 2
Why this answer is correct
The correct answer is A. \(,\sqrt{7}-\sqrt{5},\). Multiplying by \(\sqrt{7}-\sqrt{5}\) makes the denominator (7-5=2) and gives \(\sqrt{7}-\sqrt{5}\). In exams, use the conjugate.
Step 3
Exam Tip
हर को \(\sqrt{7}-\sqrt{5}\) से गुणा करने पर हर (7-5=2) होता है और उत्तर \(\sqrt{7}-\sqrt{5}\) मिलता है। परीक्षा में conjugate का प्रयोग करें।
Since \(4^{x+1}=2^{2x+2}\) and \(128=2^7\), we get (2x+2=7) and \(x=\dfrac{5}{2}\). In exams, write both sides with the same base.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{5}{2},\). Since \(4^{x+1}=2^{2x+2}\) and \(128=2^7\), we get (2x+2=7) and \(x=\dfrac{5}{2}\). In exams, write both sides with the same base.
Step 3
Exam Tip
क्योंकि \(4^{x+1}=2^{2x+2}\) और \(128=2^7\), इसलिए (2x+2=7) तथा \(x=\dfrac{5}{2}\)। परीक्षा में दोनों पक्षों को समान आधार में लिखें।
On expansion, ((2m-n)2=4m-2-4mn+n-2) and ((m+n)2=m-2+2mn+n-2), so the difference is \(3m^2-6mn\). In exams, check the signs carefully.
Step 2
Why this answer is correct
The correct answer is A. \(,3m^2-6mn,\). On expansion, ((2m-n)2=4m-2-4mn+n-2) and ((m+n)2=m-2+2mn+n-2), so the difference is \(3m^2-6mn\). In exams, check the signs carefully.
Step 3
Exam Tip
विस्तार करने पर ((2m-n)2=4m-2-4mn+n-2) और ((m+n)2=m-2+2mn+n-2), इसलिए अंतर \(3m^2-6mn\) है। परीक्षा में चिन्हों की जांच करें।
The product of coefficients (-4) and (3) is (-12), and \(a^{2-1}b^{-3+5}=ab^2\). In exams, handle coefficients and exponents separately.
Step 2
Why this answer is correct
The correct answer is A. \(,-12ab^2,\). The product of coefficients (-4) and (3) is (-12), and \(a^{2-1}b^{-3+5}=ab^2\). In exams, handle coefficients and exponents separately.
Step 3
Exam Tip
गुणांक (-4) और (3) का गुणनफल (-12) है, और \(a^{2-1}b^{-3+5}=ab^2\) है। परीक्षा में गुणांक और घातांक अलग-अलग संभालें।
Because (\(5x^2\)0=1), \(x^0=1\), and \(2^{-1}=\dfrac{1}{2}\), the value is (4). In exams, apply the zero exponent rule only to a non-zero base.
Step 2
Why this answer is correct
The correct answer is A. (,4,). Because (\(5x^2\)0=1), \(x^0=1\), and \(2^{-1}=\dfrac{1}{2}\), the value is (4). In exams, apply the zero exponent rule only to a non-zero base.
Step 3
Exam Tip
क्योंकि (\(5x^2\)0=1), \(x^0=1\) और \(2^{-1}=\dfrac{1}{2}\), इसलिए मान (4) है। परीक्षा में शून्य घात का नियम केवल non-zero आधार पर लगाएं।
(\left\(\dfrac{27}{8}\right\)^{\frac{1}{3}}=\dfrac{3}{2}), so (\left\(\dfrac{27}{8}\right\)^{-\frac{2}{3}}=\left\(\dfrac{3}{2}\right\)^{-2}=\dfrac{4}{9}). In exams, take the reciprocal for a negative exponent.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{4}{9},\). (\left\(\dfrac{27}{8}\right\)^{\frac{1}{3}}=\dfrac{3}{2}), so (\left\(\dfrac{27}{8}\right\)^{-\frac{2}{3}}=\left\(\dfrac{3}{2}\right\)^{-2}=\dfrac{4}{9}). In exams, take the reciprocal for a negative exponent.
Step 3
Exam Tip
(\left\(\dfrac{27}{8}\right\)^{\frac{1}{3}}=\dfrac{3}{2}), इसलिए (\left\(\dfrac{27}{8}\right\)^{-\frac{2}{3}}=\left\(\dfrac{3}{2}\right\)^{-2}=\dfrac{4}{9})। परीक्षा में ऋणात्मक घात में reciprocal लें।
\(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the inside value is \(-\sqrt{3}\) and the product is (-6). In exams, simplify the surds first.
Step 2
Why this answer is correct
The correct answer is A. (,-6,). \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the inside value is \(-\sqrt{3}\) and the product is (-6). In exams, simplify the surds first.
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{27}=3\sqrt{3}\), इसलिए अंदर का मान \(-\sqrt{3}\) है और गुणनफल (-6) है। परीक्षा में पहले surd को सरल करें।
This matches ((a-b)\(a^2+ab+b^2\)=a-3-b-3), so the answer is \(x^3-8\). In exams, identifying the identity makes expansion faster.
Step 2
Why this answer is correct
The correct answer is A. \(,x^3-8,\). This matches ((a-b)\(a^2+ab+b^2\)=a-3-b-3), so the answer is \(x^3-8\). In exams, identifying the identity makes expansion faster.
Step 3
Exam Tip
यह ((a-b)\(a^2+ab+b^2\)=a-3-b-3) का रूप है, इसलिए उत्तर \(x^3-8\) है। परीक्षा में identity पहचानने से विस्तार जल्दी होता है।
Applying the outside exponent \(\dfrac{1}{2}\) gives \(a^2b^{-1}=\dfrac{a^2}{b}\). In exams, apply the fractional power to every factor.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{a^2}{b},\). Applying the outside exponent \(\dfrac{1}{2}\) gives \(a^2b^{-1}=\dfrac{a^2}{b}\). In exams, apply the fractional power to every factor.
Step 3
Exam Tip
बाहर की घात \(\dfrac{1}{2}\) लगाने पर \(a^2b^{-1}=\dfrac{a^2}{b}\) मिलता है। परीक्षा में fractional power को हर factor पर लगाएं।
Taking \(10^4\) common in the numerator gives \(\dfrac{10^4(10-1)}{9\times 10^3}=10\). In exams, taking a common factor makes calculation easier.
Step 2
Why this answer is correct
The correct answer is A. (,10,). Taking \(10^4\) common in the numerator gives \(\dfrac{10^4(10-1)}{9\times 10^3}=10\). In exams, taking a common factor makes calculation easier.
Step 3
Exam Tip
ऊपर \(10^4\) common लेने पर \(\dfrac{10^4(10-1)}{9\times 10^3}=10\) मिलता है। परीक्षा में common factor लेने से गणना आसान होती है।
\(\dfrac{\sqrt{48}}{\sqrt{3}}=\sqrt{16}=4\) and \(\dfrac{\sqrt{75}}{\sqrt{3}}=\sqrt{25}=5\), so the sum is (9). In exams, simplify the division inside the root.
Step 2
Why this answer is correct
The correct answer is A. (,9,). \(\dfrac{\sqrt{48}}{\sqrt{3}}=\sqrt{16}=4\) and \(\dfrac{\sqrt{75}}{\sqrt{3}}=\sqrt{25}=5\), so the sum is (9). In exams, simplify the division inside the root.
Step 3
Exam Tip
\(\dfrac{\sqrt{48}}{\sqrt{3}}=\sqrt{16}=4\) और \(\dfrac{\sqrt{75}}{\sqrt{3}}=\sqrt{25}=5\), इसलिए योग (9) है। परीक्षा में root के अंदर भाग को सरल करें।
Since \(81=3^4\), we get (2x-1=4) and \(x=\dfrac{5}{2}\). In exams, equate exponents when the bases are the same.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{5}{2},\). Since \(81=3^4\), we get (2x-1=4) and \(x=\dfrac{5}{2}\). In exams, equate exponents when the bases are the same.
Step 3
Exam Tip
क्योंकि \(81=3^4\), इसलिए (2x-1=4) और \(x=\dfrac{5}{2}\)। परीक्षा में समान आधार होने पर घातांकों को बराबर करें।
Inside, \(\dfrac{a^2}{b^{-3}}=a^2b^3\), and applying the power (-2) gives \(\dfrac{1}{a^4b^6}\). In exams, simplify the inside part first.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{1}{a^4b^6},\). Inside, \(\dfrac{a^2}{b^{-3}}=a^2b^3\), and applying the power (-2) gives \(\dfrac{1}{a^4b^6}\). In exams, simplify the inside part first.
Step 3
Exam Tip
अंदर \(\dfrac{a^2}{b^{-3}}=a^2b^3\), और (-2) घात लगाने पर \(\dfrac{1}{a^4b^6}\) मिलता है। परीक्षा में अंदर का भाग पहले सरल करें।
Since \(\sqrt{8}=2\sqrt{2}\), (\(\sqrt{2}+\sqrt{8}\)2=\(3\sqrt{2}\)2=18). In exams, simplify the surd before squaring.
Step 2
Why this answer is correct
The correct answer is A. (,18,). Since \(\sqrt{8}=2\sqrt{2}\), (\(\sqrt{2}+\sqrt{8}\)2=\(3\sqrt{2}\)2=18). In exams, simplify the surd before squaring.
Step 3
Exam Tip
क्योंकि \(\sqrt{8}=2\sqrt{2}\), इसलिए (\(\sqrt{2}+\sqrt{8}\)2=\(3\sqrt{2}\)2=18)। परीक्षा में वर्ग करने से पहले surd सरल करें।
This is of the form ((A+B)2-(A-B)2=4AB), where (A=3x) and (B=2), so the answer is (24x). In exams, identities save time.
Step 2
Why this answer is correct
The correct answer is A. (,24x,). This is of the form ((A+B)2-(A-B)2=4AB), where (A=3x) and (B=2), so the answer is (24x). In exams, identities save time.
Step 3
Exam Tip
यह ((A+B)2-(A-B)2=4AB) का रूप है, जहां (A=3x) और (B=2), इसलिए उत्तर (24x) है। परीक्षा में identity से समय बचता है।
(\left\(\dfrac{9}{4}\right\)^{\frac{1}{2}}=\dfrac{3}{2}), so (\left\(\dfrac{9}{4}\right\)^{\frac{3}{2}}=\left\(\dfrac{3}{2}\right\)3=\dfrac{27}{8}). In exams, take the square root first.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{27}{8},\). (\left\(\dfrac{9}{4}\right\)^{\frac{1}{2}}=\dfrac{3}{2}), so (\left\(\dfrac{9}{4}\right\)^{\frac{3}{2}}=\left\(\dfrac{3}{2}\right\)3=\dfrac{27}{8}). In exams, take the square root first.
Step 3
Exam Tip
(\left\(\dfrac{9}{4}\right\)^{\frac{1}{2}}=\dfrac{3}{2}), इसलिए (\left\(\dfrac{9}{4}\right\)^{\frac{3}{2}}=\left\(\dfrac{3}{2}\right\)3=\dfrac{27}{8})। परीक्षा में square root पहले निकालें।
Inside, \(2^{-3}+2^{-2}=\dfrac{1}{8}+\dfrac{1}{4}=\dfrac{3}{8}\), so the power (-1) gives \(\dfrac{8}{3}\). In exams, simplify the bracket first.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{8}{3},\). Inside, \(2^{-3}+2^{-2}=\dfrac{1}{8}+\dfrac{1}{4}=\dfrac{3}{8}\), so the power (-1) gives \(\dfrac{8}{3}\). In exams, simplify the bracket first.
Step 3
Exam Tip
अंदर \(2^{-3}+2^{-2}=\dfrac{1}{8}+\dfrac{1}{4}=\dfrac{3}{8}\), इसलिए (-1) घात से \(\dfrac{8}{3}\) मिलता है। परीक्षा में bracket को पहले सरल करें।
(x-4-81=\(x^2-9\)\(x^2+9\)), so the simplified form is \(x^2-9\). In exams, treat \(x^4\) as (\(x^2\)2) while factoring.
Step 2
Why this answer is correct
The correct answer is A. \(,x^2-9,\). (x-4-81=\(x^2-9\)\(x^2+9\)), so the simplified form is \(x^2-9\). In exams, treat \(x^4\) as (\(x^2\)2) while factoring.
Step 3
Exam Tip
(x-4-81=\(x^2-9\)\(x^2+9\)), इसलिए सरल रूप \(x^2-9\) है। परीक्षा में \(x^4\) को (\(x^2\)2) मानकर factor करें।
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।
On expansion, ((x+1)3=x-3+3x-2+3x+1) and ((x-1)3=x-3-3x-2+3x-1), so the difference is \(6x^2+2\). In exams, expand cubes carefully.
Step 2
Why this answer is correct
The correct answer is A. \(,6x^2+2,\). On expansion, ((x+1)3=x-3+3x-2+3x+1) and ((x-1)3=x-3-3x-2+3x-1), so the difference is \(6x^2+2\). In exams, expand cubes carefully.
Step 3
Exam Tip
विस्तार करने पर ((x+1)3=x-3+3x-2+3x+1) और ((x-1)3=x-3-3x-2+3x-1), इसलिए अंतर \(6x^2+2\) है। परीक्षा में cube expansion ध्यान से करें।
((64)^{\frac{1}{3}}=4), (\(x^6\)^{\frac{1}{3}}=x-2), and (\(y^{-3}\)^{\frac{1}{3}}=y^{-1}), so the answer is \(\dfrac{4x^2}{y}\). In exams, apply the exponent to each factor.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{4x^2}{y},\). ((64)^{\frac{1}{3}}=4), (\(x^6\)^{\frac{1}{3}}=x-2), and (\(y^{-3}\)^{\frac{1}{3}}=y^{-1}), so the answer is \(\dfrac{4x^2}{y}\). In exams, apply the exponent to each factor.
Step 3
Exam Tip
((64)^{\frac{1}{3}}=4), (\(x^6\)^{\frac{1}{3}}=x-2) और (\(y^{-3}\)^{\frac{1}{3}}=y^{-1}), इसलिए उत्तर \(\dfrac{4x^2}{y}\) है। परीक्षा में प्रत्येक factor पर घात लगाएं।
Because \(\sqrt{a^4}=a^2\) and \(\sqrt{b^2}=b\), the simplified form is \(a^2b\). In exams, note the positive condition.
Step 2
Why this answer is correct
The correct answer is A. \(,a^2b,\). Because \(\sqrt{a^4}=a^2\) and \(\sqrt{b^2}=b\), the simplified form is \(a^2b\). In exams, note the positive condition.
Step 3
Exam Tip
क्योंकि \(\sqrt{a^4}=a^2\) और \(\sqrt{b^2}=b\), इसलिए सरल रूप \(a^2b\) है। परीक्षा में positive condition को ध्यान में रखें।
The numerator difference is (6x-2y+2y-3=2y\(3x^2+y^2\)), so division gives \(3x^2+y^2\). In exams, take out the common factor.
Step 2
Why this answer is correct
The correct answer is A. \(,3x^2+y^2,\). The numerator difference is (6x-2y+2y-3=2y\(3x^2+y^2\)), so division gives \(3x^2+y^2\). In exams, take out the common factor.
Step 3
Exam Tip
ऊपर का अंतर (6x-2y+2y-3=2y\(3x^2+y^2\)) है, इसलिए भाग देने पर \(3x^2+y^2\) मिलता है। परीक्षा में common factor निकालें।
\(9^2=3^4\) and \(27^{-1}=3^{-3}\), so the value is \(3^{-2+4-(-3)}=3^5=243\). In exams, be careful while subtracting a negative exponent.
Step 2
Why this answer is correct
The correct answer is A. (,243,). \(9^2=3^4\) and \(27^{-1}=3^{-3}\), so the value is \(3^{-2+4-(-3)}=3^5=243\). In exams, be careful while subtracting a negative exponent.
Step 3
Exam Tip
\(9^2=3^4\) और \(27^{-1}=3^{-3}\), इसलिए मान \(3^{-2+4-(-3)}=3^5=243\) है। परीक्षा में negative exponent घटाते समय सावधान रहें।
The numerator is ((2x)3\(3x^{-2}\)=8x-3\cdot 3x^{-2}=24x), and \(\dfrac{24x}{12x^{-1}}=2x^2\). In exams, simplify both coefficient and variable parts.
Step 2
Why this answer is correct
The correct answer is A. \(,2x^2,\). The numerator is ((2x)3\(3x^{-2}\)=8x-3\cdot 3x^{-2}=24x), and \(\dfrac{24x}{12x^{-1}}=2x^2\). In exams, simplify both coefficient and variable parts.
Step 3
Exam Tip
ऊपर ((2x)3\(3x^{-2}\)=8x-3\cdot 3x^{-2}=24x), और \(\dfrac{24x}{12x^{-1}}=2x^2\)। परीक्षा में coefficient और variable दोनों सरल करें।
\(\sqrt{98}=7\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), and \(\sqrt{50}=5\sqrt{2}\), so the answer is \(8\sqrt{2}\). In exams, first write all surds in simplest form.
Step 2
Why this answer is correct
The correct answer is A. \(,8\sqrt{2},\). \(\sqrt{98}=7\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), and \(\sqrt{50}=5\sqrt{2}\), so the answer is \(8\sqrt{2}\). In exams, first write all surds in simplest form.
Step 3
Exam Tip
\(\sqrt{98}=7\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\) और \(\sqrt{50}=5\sqrt{2}\), इसलिए उत्तर \(8\sqrt{2}\) है। परीक्षा में पहले सभी surds को simplest form में लिखें।
Multiplying by \(2+\sqrt{3}\) makes the denominator (4-3=1). In exams, multiply both numerator and denominator by the conjugate.
Step 2
Why this answer is correct
The correct answer is A. \(,6+3\sqrt{3},\). Multiplying by \(2+\sqrt{3}\) makes the denominator (4-3=1). In exams, multiply both numerator and denominator by the conjugate.
Step 3
Exam Tip
हर को \(2+\sqrt{3}\) से गुणा करने पर हर (4-3=1) हो जाता है। परीक्षा में conjugate से numerator और denominator दोनों को गुणा करें।
\(4^{-1}-5^{-1}=\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{1}{20}\), so the whole value is (20). In exams, first convert negative powers into fractions.
Step 2
Why this answer is correct
The correct answer is A. (,20,). \(4^{-1}-5^{-1}=\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{1}{20}\), so the whole value is (20). In exams, first convert negative powers into fractions.
Step 3
Exam Tip
\(4^{-1}-5^{-1}=\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{1}{20}\), इसलिए पूरा मान (20) है। परीक्षा में negative powers को पहले fractions में बदलें।
\(125^{\frac{2}{3}}=25\) and \(25^{\frac{1}{2}}=5\), so the value is (5). In exams, separate fractional exponents into root and power.
Step 2
Why this answer is correct
The correct answer is A. (,5,). \(125^{\frac{2}{3}}=25\) and \(25^{\frac{1}{2}}=5\), so the value is (5). In exams, separate fractional exponents into root and power.
Step 3
Exam Tip
\(125^{\frac{2}{3}}=25\) और \(25^{\frac{1}{2}}=5\), इसलिए मान (5) है। परीक्षा में fractional exponents को root और power में अलग करें।
The numerator is \(a^{-1}+b^{-1}=\dfrac{a+b}{ab}\) and the denominator is ((ab)^{-1}=\dfrac{1}{ab}), so the answer is (a+b). In exams, make a common denominator.
Step 2
Why this answer is correct
The correct answer is A. (,a+b,). The numerator is \(a^{-1}+b^{-1}=\dfrac{a+b}{ab}\) and the denominator is ((ab)^{-1}=\dfrac{1}{ab}), so the answer is (a+b). In exams, make a common denominator.
Step 3
Exam Tip
ऊपर \(a^{-1}+b^{-1}=\dfrac{a+b}{ab}\) और नीचे ((ab)^{-1}=\dfrac{1}{ab}), इसलिए उत्तर (a+b) है। परीक्षा में common denominator बनाएं।
(\(2^5\)^{\frac{2}{5}}=22=4) and (\(3^3\)^{\frac{1}{3}}=3), so the product is (12). In exams, apply the power of a power law.
Step 2
Why this answer is correct
The correct answer is A. (,12,). (\(2^5\)^{\frac{2}{5}}=22=4) and (\(3^3\)^{\frac{1}{3}}=3), so the product is (12). In exams, apply the power of a power law.
Step 3
Exam Tip
(\(2^5\)^{\frac{2}{5}}=22=4) और (\(3^3\)^{\frac{1}{3}}=3), इसलिए गुणनफल (12) है। परीक्षा में power of power नियम लगाएं।
