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।
\(\dfrac{\sqrt{48}}{\sqrt{3}}=\sqrt{16}=4\) 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।
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।
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।
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।
\(\sqrt{12}=2\sqrt{3}\) 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।
(\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।
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।
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।
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।
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।
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।
\(\sqrt{75}=5\sqrt{3}\), \(\sqrt{12}=2\sqrt{3}\), 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।
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।
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।
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।
\(2^{-1}+3^{-1}=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\), so the whole value is \(\dfrac{6}{5}\). In exams, simplify the denominator first.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{6}{5},\). \(2^{-1}+3^{-1}=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\), so the whole value is \(\dfrac{6}{5}\). In exams, simplify the denominator first.
Step 3
Exam Tip
\(2^{-1}+3^{-1}=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\), इसलिए पूरा मान \(\dfrac{6}{5}\) है। परीक्षा में denominator को पहले simplify करें।
Since \(\sqrt[3]{64}=4\), \(4^{-2}=\dfrac{1}{16}\). In exams, first evaluate the root and then apply the negative exponent.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{1}{16},\). Since \(\sqrt[3]{64}=4\), \(4^{-2}=\dfrac{1}{16}\). In exams, first evaluate the root and then apply the negative exponent.
Step 3
Exam Tip
क्योंकि \(\sqrt[3]{64}=4\), इसलिए \(4^{-2}=\dfrac{1}{16}\)। परीक्षा में पहले root का मान निकालें फिर negative exponent लगाएं।
The expression inside is \(a^{-3}b^4\), and the power (-1) gives its reciprocal \(\dfrac{a^3}{b^4}\). In exams, apply the outer negative power at the end.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{a^3}{b^4},\). The expression inside is \(a^{-3}b^4\), and the power (-1) gives its reciprocal \(\dfrac{a^3}{b^4}\). In exams, apply the outer negative power at the end.
Step 3
Exam Tip
अंदर का भाग \(a^{-3}b^4\) है, और (-1) घात से उसका व्युत्क्रम \(\dfrac{a^3}{b^4}\) हो जाता है। परीक्षा में outer negative power अंत में लगाएं।
\(x^{5-(-1)}=x^6\) and \(y^{-2-3}=y^{-5}\), so the form is \(\dfrac{x^6}{y^5}\). In exams, simplify the exponent of each variable separately.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{x^6}{y^5},\). \(x^{5-(-1)}=x^6\) and \(y^{-2-3}=y^{-5}\), so the form is \(\dfrac{x^6}{y^5}\). In exams, simplify the exponent of each variable separately.
Step 3
Exam Tip
\(x^{5-(-1)}=x^6\) और \(y^{-2-3}=y^{-5}\), इसलिए रूप \(\dfrac{x^6}{y^5}\) है। परीक्षा में हर variable का exponent अलग-अलग simplify करें।
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।
When both expansions are added, (2mn) and (-2mn) cancel, giving \(2m^2+2n^2\). In exams, notice opposite middle terms.
Step 2
Why this answer is correct
The correct answer is A. \(,2m^2+2n^2,\). When both expansions are added, (2mn) and (-2mn) cancel, giving \(2m^2+2n^2\). In exams, notice opposite middle terms.
Step 3
Exam Tip
दोनों विस्तारों को जोड़ने पर (2mn) और (-2mn) कट जाते हैं, इसलिए \(2m^2+2n^2\) मिलता है। परीक्षा में opposite middle terms पर ध्यान दें।
