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Here \(5^0=1\), \(3^{-1}=\dfrac{1}{3}\), and \(2^{-2}=\dfrac{1}{4}\), so the value is \(\dfrac{16}{3}\). In exams, first convert negative exponents into fractions.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{16}{3},\). Here \(5^0=1\), \(3^{-1}=\dfrac{1}{3}\), and \(2^{-2}=\dfrac{1}{4}\), so the value is \(\dfrac{16}{3}\). In exams, first convert negative exponents into fractions.
Step 3
Exam Tip
यहां \(5^0=1\), \(3^{-1}=\dfrac{1}{3}\) और \(2^{-2}=\dfrac{1}{4}\), इसलिए मान \(\dfrac{16}{3}\) है। परीक्षा में ऋणात्मक घात को पहले भिन्न में बदलें।
The numerator gives \(a^m \times a^{2m}=a^{3m}\), and then \(\dfrac{a^{3m}}{a^{3m-2}}=a^2\). In exams, subtract exponents during division.
Step 2
Why this answer is correct
The correct answer is A. \(,a^2,\). The numerator gives \(a^m \times a^{2m}=a^{3m}\), and then \(\dfrac{a^{3m}}{a^{3m-2}}=a^2\). In exams, subtract exponents during division.
Step 3
Exam Tip
ऊपर \(a^m \times a^{2m}=a^{3m}\) और फिर \(\dfrac{a^{3m}}{a^{3m-2}}=a^2\) होगा। परीक्षा में भाग करते समय घातांक घटाएं।
The outside power (-2) multiplies both exponents, so \(x^4y^{-6}=\dfrac{x^4}{y^6}\). In exams, apply the outside power to every factor inside the bracket.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{x^4}{y^6},\). The outside power (-2) multiplies both exponents, so \(x^4y^{-6}=\dfrac{x^4}{y^6}\). In exams, apply the outside power to every factor inside the bracket.
Step 3
Exam Tip
बाहर की घात (-2) दोनों घातांकों से गुणा होगी, इसलिए \(x^4y^{-6}=\dfrac{x^4}{y^6}\) है। परीक्षा में bracket के बाहर की घात को हर factor पर लगाएं।
Here \(16^{\frac{1}{4}}=2\), so \(16^{\frac{3}{4}}=8\) and \(16^{-\frac{3}{4}}=\dfrac{1}{8}\). In exams, a negative exponent means reciprocal.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{1}{8},\). Here \(16^{\frac{1}{4}}=2\), so \(16^{\frac{3}{4}}=8\) and \(16^{-\frac{3}{4}}=\dfrac{1}{8}\). In exams, a negative exponent means reciprocal.
Step 3
Exam Tip
यहां \(16^{\frac{1}{4}}=2\), इसलिए \(16^{\frac{3}{4}}=8\) और \(16^{-\frac{3}{4}}=\dfrac{1}{8}\)। परीक्षा में ऋणात्मक घात का अर्थ व्युत्क्रम होता है।
Because \(\sqrt{50}=5\sqrt{2}\), \(\sqrt{8}=2\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\), the answer is \(4\sqrt{2}\). In exams, combine only like surd terms.
Step 2
Why this answer is correct
The correct answer is A. \(,4\sqrt{2},\). Because \(\sqrt{50}=5\sqrt{2}\), \(\sqrt{8}=2\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\), the answer is \(4\sqrt{2}\). In exams, combine only like surd terms.
Step 3
Exam Tip
क्योंकि \(\sqrt{50}=5\sqrt{2}\), \(\sqrt{8}=2\sqrt{2}\) और \(\sqrt{18}=3\sqrt{2}\), इसलिए उत्तर \(4\sqrt{2}\) है। परीक्षा में समान surd terms को ही जोड़ें या घटाएं।
Multiplying by \(\sqrt{3}+\sqrt{2}\) makes the denominator (3-2=1). In exams, remember to multiply by the conjugate.
Step 2
Why this answer is correct
The correct answer is A. \(,\sqrt{3}+\sqrt{2},\). Multiplying by \(\sqrt{3}+\sqrt{2}\) makes the denominator (3-2=1). In exams, remember to multiply by the conjugate.
Step 3
Exam Tip
हर को \(\sqrt{3}+\sqrt{2}\) से गुणा करने पर हर (3-2=1) हो जाता है। परीक्षा में conjugate से गुणा करना न भूलें।
(\left\(\dfrac{2}{3}\right\)^{-2}=\left\(\dfrac{3}{2}\right\)2=\dfrac{9}{4}), so the product is (1). In exams, a fraction is inverted under a negative exponent.
Step 2
Why this answer is correct
The correct answer is A. (,1,). (\left\(\dfrac{2}{3}\right\)^{-2}=\left\(\dfrac{3}{2}\right\)2=\dfrac{9}{4}), so the product is (1). In exams, a fraction is inverted under a negative exponent.
Step 3
Exam Tip
(\left\(\dfrac{2}{3}\right\)^{-2}=\left\(\dfrac{3}{2}\right\)2=\dfrac{9}{4}), इसलिए गुणनफल (1) है। परीक्षा में ऋणात्मक घात में भिन्न उलट जाती है।
Here \(27^{\frac{2}{3}}=9\) and \(81^{\frac{1}{4}}=3\), so the product is (27). In exams, first take the root and then apply the power.
Step 2
Why this answer is correct
The correct answer is A. (,27,). Here \(27^{\frac{2}{3}}=9\) and \(81^{\frac{1}{4}}=3\), so the product is (27). In exams, first take the root and then apply the power.
Step 3
Exam Tip
यहां \(27^{\frac{2}{3}}=9\) और \(81^{\frac{1}{4}}=3\), इसलिए गुणनफल (27) है। परीक्षा में पहले मूल निकालें फिर घात लगाएं।
From \(9=3^2\), (a=2), and from \(8=2^3\), (b=3), so (a+b=5). In exams, remembering small powers gives faster solutions.
Step 2
Why this answer is correct
The correct answer is A. (,5,). From \(9=3^2\), (a=2), and from \(8=2^3\), (b=3), so (a+b=5). In exams, remembering small powers gives faster solutions.
Step 3
Exam Tip
\(9=3^2\) से (a=2) और \(8=2^3\) से (b=3), इसलिए (a+b=5)। परीक्षा में छोटे powers को याद रखना तेज समाधान देता है।
Inside, \(a^{\frac{1}{2}}a^{\frac{3}{2}}=a^2\), so (\dfrac{\(a^2\)2}{a-3}=a). In exams, solve fractional exponents using the usual exponent rules.
Step 2
Why this answer is correct
The correct answer is A. (,a,). Inside, \(a^{\frac{1}{2}}a^{\frac{3}{2}}=a^2\), so (\dfrac{\(a^2\)2}{a-3}=a). In exams, solve fractional exponents using the usual exponent rules.
Step 3
Exam Tip
अंदर \(a^{\frac{1}{2}}a^{\frac{3}{2}}=a^2\), इसलिए (\dfrac{\(a^2\)2}{a-3}=a)। परीक्षा में fractional exponents को भी सामान्य घात नियम से हल करें।
Because (x-2-y-2=(x-y)(x+y)), the simplified form is (x+y). In exams, identifying difference of squares is very useful.
Step 2
Why this answer is correct
The correct answer is A. (,x+y,). Because (x-2-y-2=(x-y)(x+y)), the simplified form is (x+y). In exams, identifying difference of squares is very useful.
Step 3
Exam Tip
क्योंकि (x-2-y-2=(x-y)(x+y)), इसलिए सरल रूप (x+y) है। परीक्षा में difference of squares पहचानना बहुत उपयोगी है।
On expansion, ((p+q)2=p-2+2pq+q-2) and ((p-q)2=p-2-2pq+q-2), so the difference is (4pq). In exams, apply standard identities directly.
Step 2
Why this answer is correct
The correct answer is A. (,4pq,). On expansion, ((p+q)2=p-2+2pq+q-2) and ((p-q)2=p-2-2pq+q-2), so the difference is (4pq). In exams, apply standard identities directly.
Step 3
Exam Tip
विस्तार करने पर ((p+q)2=p-2+2pq+q-2) और ((p-q)2=p-2-2pq+q-2), इसलिए अंतर (4pq) है। परीक्षा में standard identities सीधे लगाएं।
The product of coefficients (2) and (-3) is (-6), and powers of like variables are added. In exams, watch both the sign and the exponents carefully.
Step 2
Why this answer is correct
The correct answer is A. \(,-6x^3y^3,\). The product of coefficients (2) and (-3) is (-6), and powers of like variables are added. In exams, watch both the sign and the exponents carefully.
Step 3
Exam Tip
गुणांक (2) और (-3) का गुणनफल (-6) है, और समान चरों की घातें जुड़ती हैं। परीक्षा में sign और exponents दोनों ध्यान से देखें।
The coefficient is \(\dfrac{6}{2}=3\), \(a^{3-1}=a^2\), and \(b^{2-(-1)}=b^3\). In exams, the sign changes when subtracting a negative exponent.
Step 2
Why this answer is correct
The correct answer is A. \(,3a^2b^3,\). The coefficient is \(\dfrac{6}{2}=3\), \(a^{3-1}=a^2\), and \(b^{2-(-1)}=b^3\). In exams, the sign changes when subtracting a negative exponent.
Step 3
Exam Tip
गुणांक \(\dfrac{6}{2}=3\), \(a^{3-1}=a^2\) और \(b^{2-(-1)}=b^3\) है। परीक्षा में हर के ऋणात्मक घातांक को घटाते समय sign बदलता है।
The numerator is (\(x^3\)2=x-6) and the denominator is \(x^{-1}x^4=x^3\), so the answer is \(x^3\). In exams, apply an exponent law at each step.
Step 2
Why this answer is correct
The correct answer is A. \(,x^3,\). The numerator is (\(x^3\)2=x-6) and the denominator is \(x^{-1}x^4=x^3\), so the answer is \(x^3\). In exams, apply an exponent law at each step.
Step 3
Exam Tip
ऊपर (\(x^3\)2=x-6) और नीचे \(x^{-1}x^4=x^3\), इसलिए उत्तर \(x^3\) है। परीक्षा में हर step पर exponent law अलग से लगाएं।
Since \(0.00032=3.2\times 10^{-4}\), \(\dfrac{3.2\times 10^{-4}}{10^{-5}}=3.2\times 10^1=32\). In exams, converting decimals to scientific notation helps.
Step 2
Why this answer is correct
The correct answer is A. (,32,). Since \(0.00032=3.2\times 10^{-4}\), \(\dfrac{3.2\times 10^{-4}}{10^{-5}}=3.2\times 10^1=32\). In exams, converting decimals to scientific notation helps.
Step 3
Exam Tip
क्योंकि \(0.00032=3.2\times 10^{-4}\), इसलिए \(\dfrac{3.2\times 10^{-4}}{10^{-5}}=3.2\times 10^1=32\)। परीक्षा में decimal को scientific notation में बदलना मदद करता है।
Taking \(7^4\) common in the numerator gives (\dfrac{74(7-1)}{74}=6). In exams, taking a common factor makes calculation shorter.
Step 2
Why this answer is correct
The correct answer is A. (,6,). Taking \(7^4\) common in the numerator gives (\dfrac{74(7-1)}{74}=6). In exams, taking a common factor makes calculation shorter.
Step 3
Exam Tip
ऊपर से \(7^4\) common लेने पर (\dfrac{74(7-1)}{74}=6) मिलता है। परीक्षा में समान factor common लेना गणना को छोटा करता है।
The numerator is \(2^{10}+2^{10}=2\times 2^{10}=2^{11}\), so \(\dfrac{2^{11}}{2^9}=2^2=4\). In exams, first combine like terms and then apply exponent laws.
Step 2
Why this answer is correct
The correct answer is A. (,4,). The numerator is \(2^{10}+2^{10}=2\times 2^{10}=2^{11}\), so \(\dfrac{2^{11}}{2^9}=2^2=4\). In exams, first combine like terms and then apply exponent laws.
Step 3
Exam Tip
ऊपर \(2^{10}+2^{10}=2\times 2^{10}=2^{11}\), इसलिए \(\dfrac{2^{11}}{2^9}=2^2=4\)। परीक्षा में पहले समान terms को जोड़ें फिर घात नियम लगाएं।
Because \(\sqrt{a^2}=a\) and \(\sqrt{b^4}=b^2\), the answer is \(ab^2\). In exams, note the condition that variables are positive.
Step 2
Why this answer is correct
The correct answer is A. \(,ab^2,\). Because \(\sqrt{a^2}=a\) and \(\sqrt{b^4}=b^2\), the answer is \(ab^2\). In exams, note the condition that variables are positive.
Step 3
Exam Tip
क्योंकि \(\sqrt{a^2}=a\) और \(\sqrt{b^4}=b^2\), इसलिए उत्तर \(ab^2\) है। परीक्षा में variables के positive होने की शर्त ध्यान रखें।
\(\dfrac{x^2}{y^{-1}}=x^2y\), so the whole square is \(x^4y^2\). In exams, simplify a negative exponent by moving its position.
Step 2
Why this answer is correct
The correct answer is A. \(,x^4y^2,\). \(\dfrac{x^2}{y^{-1}}=x^2y\), so the whole square is \(x^4y^2\). In exams, simplify a negative exponent by moving its position.
Step 3
Exam Tip
\(\dfrac{x^2}{y^{-1}}=x^2y\), इसलिए पूरा वर्ग \(x^4y^2\) है। परीक्षा में ऋणात्मक घातांक को स्थान बदलकर सरल करें।
A negative exponent inverts the fraction, so (\left\(-\dfrac{1}{2}\right\)^{-3}=(-2)3=-8). In exams, keep the sign of a negative base according to the power.
Step 2
Why this answer is correct
The correct answer is A. (,-8,). A negative exponent inverts the fraction, so (\left\(-\dfrac{1}{2}\right\)^{-3}=(-2)3=-8). In exams, keep the sign of a negative base according to the power.
Step 3
Exam Tip
ऋणात्मक घात से भिन्न उलटती है, इसलिए (\left\(-\dfrac{1}{2}\right\)^{-3}=(-2)3=-8)। परीक्षा में negative base का sign power के अनुसार रखें।
(\(4x^{-2}\)^{-1}=4^{-1}x-2=\dfrac{x-2}{4}). In exams, apply the outside exponent to every factor of a product.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{x^2}{4},\). (\(4x^{-2}\)^{-1}=4^{-1}x-2=\dfrac{x-2}{4}). In exams, apply the outside exponent to every factor of a product.
Step 3
Exam Tip
(\(4x^{-2}\)^{-1}=4^{-1}x-2=\dfrac{x-2}{4})। परीक्षा में product के हर factor पर बाहर की घात लगाएं।
When the two squares are added, the surd terms cancel and (7+7=14). In exams, irrational terms often cancel in conjugate expressions.
Step 2
Why this answer is correct
The correct answer is A. (,14,). When the two squares are added, the surd terms cancel and (7+7=14). In exams, irrational terms often cancel in conjugate expressions.
Step 3
Exam Tip
दोनों वर्ग जोड़ने पर surd terms कट जाते हैं और (7+7=14) मिलता है। परीक्षा में conjugate expressions में irrational terms अक्सर cancel होते हैं।
In division, \(a^{2-(-1)}=a^3\) and \(b^{-3-1}=b^{-4}\), so the answer is \(\dfrac{a^3}{b^4}\). In exams, subtract exponents of like variables separately.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{a^3}{b^4},\). In division, \(a^{2-(-1)}=a^3\) and \(b^{-3-1}=b^{-4}\), so the answer is \(\dfrac{a^3}{b^4}\). In exams, subtract exponents of like variables separately.
Step 3
Exam Tip
भाग में \(a^{2-(-1)}=a^3\) और \(b^{-3-1}=b^{-4}\), इसलिए उत्तर \(\dfrac{a^3}{b^4}\) है। परीक्षा में समान variables के exponents अलग-अलग घटाएं।
First, (\left\(\dfrac{81}{16}\right\)^{\frac{1}{2}}=\dfrac{9}{4}), then the negative exponent gives \(\dfrac{4}{9}\). In exams, check both the square root and the reciprocal.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{4}{9},\). First, (\left\(\dfrac{81}{16}\right\)^{\frac{1}{2}}=\dfrac{9}{4}), then the negative exponent gives \(\dfrac{4}{9}\). In exams, check both the square root and the reciprocal.
Step 3
Exam Tip
पहले (\left\(\dfrac{81}{16}\right\)^{\frac{1}{2}}=\dfrac{9}{4}), फिर ऋणात्मक घात से उत्तर \(\dfrac{4}{9}\) होता है। परीक्षा में square root और reciprocal दोनों देखें।
(\(10^3\)2=106) and \(\dfrac{10^6}{10^{-2}}=10^{6-(-2)}=10^8\). In exams, be careful while subtracting a negative exponent.
Step 2
Why this answer is correct
The correct answer is A. \(,10^8,\). (\(10^3\)2=106) and \(\dfrac{10^6}{10^{-2}}=10^{6-(-2)}=10^8\). In exams, be careful while subtracting a negative exponent.
Step 3
Exam Tip
(\(10^3\)2=106) और \(\dfrac{10^6}{10^{-2}}=10^{6-(-2)}=10^8\)। परीक्षा में negative exponent को घटाते समय सावधान रहें।
Because \(a^{-2}=\dfrac{1}{a^2}=\dfrac{1}{5}\), \(\dfrac{1}{5}+5=\dfrac{26}{5}\). In exams, write \(a^{-2}\) as \(\dfrac{1}{a^2}\).
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{26}{5},\). Because \(a^{-2}=\dfrac{1}{a^2}=\dfrac{1}{5}\), \(\dfrac{1}{5}+5=\dfrac{26}{5}\). In exams, write \(a^{-2}\) as \(\dfrac{1}{a^2}\).
Step 3
Exam Tip
क्योंकि \(a^{-2}=\dfrac{1}{a^2}=\dfrac{1}{5}\), इसलिए \(\dfrac{1}{5}+5=\dfrac{26}{5}\)। परीक्षा में \(a^{-2}\) को \(\dfrac{1}{a^2}\) लिखें।
(x-4-16=\(x^2-4\)\(x^2+4\)), so the simplified form is \(x^2+4\). In exams, treat \(x^4\) as (\(x^2\)2) for factorisation.
Step 2
Why this answer is correct
The correct answer is A. \(,x^2+4,\). (x-4-16=\(x^2-4\)\(x^2+4\)), so the simplified form is \(x^2+4\). In exams, treat \(x^4\) as (\(x^2\)2) for factorisation.
Step 3
Exam Tip
(x-4-16=\(x^2-4\)\(x^2+4\)), इसलिए सरल रूप \(x^2+4\) है। परीक्षा में \(x^4\) को (\(x^2\)2) समझकर factor करें।
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।
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।
\(x^{5-(-1)}=x^6\) 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।
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।
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।
\(2^{-1}+3^{-1}=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\), 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।
