Inside, \(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}=\frac{1}{7}r^{-5}s^{6}\). Raising to (-1) gives \(7r^{5}s^{-6}\).
Step 2
Why this answer is correct
The correct answer is A. \(7r^{5}s^{-6}\). Inside, \(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}=\frac{1}{7}r^{-5}s^{6}\). Raising to (-1) gives \(7r^{5}s^{-6}\).
Step 3
Exam Tip
अंदर \(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}=\frac{1}{7}r^{-5}s^{6}\) है। (-1) घात लेने पर \(7r^{5}s^{-6}\) मिलता है।
Inside, \(\frac{6x^{-2}y^{3}}{3x^{4}y^{-1}}=2x^{-6}y^{4}\), and its square is \(4x^{-12}y^{8}\). Multiplying by \(\frac{x^{12}}{4y^{8}}\) gives (1).
Step 2
Why this answer is correct
The correct answer is A. (1). Inside, \(\frac{6x^{-2}y^{3}}{3x^{4}y^{-1}}=2x^{-6}y^{4}\), and its square is \(4x^{-12}y^{8}\). Multiplying by \(\frac{x^{12}}{4y^{8}}\) gives (1).
Step 3
Exam Tip
अंदर \(\frac{6x^{-2}y^{3}}{3x^{4}y^{-1}}=2x^{-6}y^{4}\), इसका वर्ग \(4x^{-12}y^{8}\) है। फिर \(\frac{x^{12}}{4y^{8}}\) से गुणा करने पर (1) मिलता है।
Inside, \(\frac{2x^{-3}}{x^{2}}=2x^{-5}\), so (\left\(2x^{-5}\right\)^{-2}x^{-4}=\frac{x^{10}}{4}x^{-4}=\frac{x^{6}}{4}). In exams, subtract the inner exponents first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{x^{6}}{4}\). Inside, \(\frac{2x^{-3}}{x^{2}}=2x^{-5}\), so (\left\(2x^{-5}\right\)^{-2}x^{-4}=\frac{x^{10}}{4}x^{-4}=\frac{x^{6}}{4}). In exams, subtract the inner exponents first.
Step 3
Exam Tip
अंदर \(\frac{2x^{-3}}{x^{2}}=2x^{-5}\) है, इसलिए (\left\(2x^{-5}\right\)^{-2}x^{-4}=\frac{x^{10}}{4}x^{-4}=\frac{x^{6}}{4})। परीक्षा में पहले अंदर की घातें घटाएं।
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 (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।
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 (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 (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 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।
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।
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।
(\left\(81x^{4}\right\)^{\frac{1}{2}}=\sqrt{81x^{4}}=9x^{2}). In exams, the exponent becomes half under a square root.
Step 2
Why this answer is correct
The correct answer is A. \(9x^{2}\). (\left\(81x^{4}\right\)^{\frac{1}{2}}=\sqrt{81x^{4}}=9x^{2}). In exams, the exponent becomes half under a square root.
Step 3
Exam Tip
(\left\(81x^{4}\right\)^{\frac{1}{2}}=\sqrt{81x^{4}}=9x^{2})। परीक्षा में वर्गमूल में घात आधी हो जाती है।
(\(a^{2}b^{-1}\)^{-3}=a^{-6}b^{3}), then \(\frac{a^{-6}b^{3}}{a^{-4}b^{2}}=a^{-2}b\). In exams, subtract powers of the same base during division.
Step 2
Why this answer is correct
The correct answer is A. \(a^{-2}b\). (\(a^{2}b^{-1}\)^{-3}=a^{-6}b^{3}), then \(\frac{a^{-6}b^{3}}{a^{-4}b^{2}}=a^{-2}b\). In exams, subtract powers of the same base during division.
Step 3
Exam Tip
(\(a^{2}b^{-1}\)^{-3}=a^{-6}b^{3}), फिर \(\frac{a^{-6}b^{3}}{a^{-4}b^{2}}=a^{-2}b\)। परीक्षा में भाग करते समय समान आधार की घात घटाएं।
Here (x^{6}=\(x^{2}\)^{3}=27) and (x^{4}=\(x^{2}\)^{2}=9), so the difference is (18). In exams, express powers using the given \(x^{2}\).
Step 2
Why this answer is correct
The correct answer is A. (18). Here (x^{6}=\(x^{2}\)^{3}=27) and (x^{4}=\(x^{2}\)^{2}=9), so the difference is (18). In exams, express powers using the given \(x^{2}\).
Step 3
Exam Tip
(x^{6}=\(x^{2}\)^{3}=27) और (x^{4}=\(x^{2}\)^{2}=9), इसलिए अंतर (18) है। परीक्षा में दी हुई घात \(x^{2}\) के रूप में लिखें।
The left side has exponents (2n) and (3n), so (2n=8) and (3n=12), giving (n=4). In exams, match exponents of both variables.
Step 2
Why this answer is correct
The correct answer is C. (4). The left side has exponents (2n) and (3n), so (2n=8) and (3n=12), giving (n=4). In exams, match exponents of both variables.
Step 3
Exam Tip
बाएँ पक्ष में घातें (2n) और (3n) हैं, इसलिए (2n=8) और (3n=12) से (n=4)। परीक्षा में दोनों चर की घात मिलाकर जांचें।
Inside, \(a^{3-(-1)}b^{-2-2}=a^{4}b^{-4}\), and squaring gives \(a^{8}b^{-8}\). In exams, watch the sign when subtracting negative exponents.
Step 2
Why this answer is correct
The correct answer is A. \(a^{8}b^{-8}\). Inside, \(a^{3-(-1)}b^{-2-2}=a^{4}b^{-4}\), and squaring gives \(a^{8}b^{-8}\). In exams, watch the sign when subtracting negative exponents.
Step 3
Exam Tip
अंदर \(a^{3-(-1)}b^{-2-2}=a^{4}b^{-4}\), इसलिए वर्ग करने पर \(a^{8}b^{-8}\) है। परीक्षा में ऋणात्मक घात घटाते समय चिह्न पर ध्यान दें।
Inside, \(\frac{x^{-2}y^{3}}{x^{4}y^{-1}}=x^{-6}y^{4}\), and raising to (-1) gives \(x^{6}y^{-4}\). In exams, subtract exponents during division.
Step 2
Why this answer is correct
The correct answer is A. \(x^{6}y^{-4}\). Inside, \(\frac{x^{-2}y^{3}}{x^{4}y^{-1}}=x^{-6}y^{4}\), and raising to (-1) gives \(x^{6}y^{-4}\). In exams, subtract exponents during division.
Step 3
Exam Tip
अंदर \(\frac{x^{-2}y^{3}}{x^{4}y^{-1}}=x^{-6}y^{4}\), और (-1) घात लेने पर \(x^{6}y^{-4}\) मिलता है। परीक्षा में भाग में घात घटती है।
Inside, \(\frac{4x^{2}y^{-3}}{2x^{-1}y}=2x^{3}y^{-4}\), and raising to (-2) gives \(\frac{y^{8}}{4x^{6}}\). In exams, simplify inside the bracket first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{y^{8}}{4x^{6}}\). Inside, \(\frac{4x^{2}y^{-3}}{2x^{-1}y}=2x^{3}y^{-4}\), and raising to (-2) gives \(\frac{y^{8}}{4x^{6}}\). In exams, simplify inside the bracket first.
Step 3
Exam Tip
अंदर \(\frac{4x^{2}y^{-3}}{2x^{-1}y}=2x^{3}y^{-4}\), इसलिए घात (-2) देने पर \(\frac{y^{8}}{4x^{6}}\) मिलता है। परीक्षा में पहले कोष्ठक के अंदर सरल करें।
Since \(25^{-2}=5^{-4}\) and \(125=5^{3}\), \(\frac{5^{8}\cdot5^{-4}}{5^{3}}=5^{1}=5\). In exams, convert (25) and (125) into powers of (5).
Step 2
Why this answer is correct
The correct answer is B. (5). Since \(25^{-2}=5^{-4}\) and \(125=5^{3}\), \(\frac{5^{8}\cdot5^{-4}}{5^{3}}=5^{1}=5\). In exams, convert (25) and (125) into powers of (5).
Step 3
Exam Tip
\(25^{-2}=5^{-4}\) और \(125=5^{3}\), इसलिए \(\frac{5^{8}\cdot5^{-4}}{5^{3}}=5^{1}=5\)। परीक्षा में (25) और (125) को (5) की घात में बदलें।
We have (\left\(3^{x}\right\)^{2}=3^{2x}) and \(729=3^{6}\), so (2x=6) and (x=3). In exams, rewrite both sides with the same base.
Step 2
Why this answer is correct
The correct answer is B. (3). We have (\left\(3^{x}\right\)^{2}=3^{2x}) and \(729=3^{6}\), so (2x=6) and (x=3). In exams, rewrite both sides with the same base.
Step 3
Exam Tip
(\left\(3^{x}\right\)^{2}=3^{2x}) और \(729=3^{6}\), इसलिए (2x=6) और (x=3)। परीक्षा में दोनों पक्षों को समान आधार में लिखें।
