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27 results found for "monomials" in Class 10.

यदि (\left\(3x^{-2}y^{3}\right\)^{2}\cdot\left\(9x^{4}y^{-1}\right\)^{-1}) को \(cx^{r}y^{s}\) लिखा जाए, तो (c+r+s) का मान क्या है?

If (\left\(3x^{-2}y^{3}\right\)^{2}\cdot\left\(9x^{4}y^{-1}\right\)^{-1}) is written as \(cx^{r}y^{s}\), what is the value of (c+r+s)?

Explanation opens after your attempt
Correct Answer

B. (2)

Step 1

Concept

The expression is \(9x^{-4}y^{6}\cdot\frac{1}{9}x^{-4}y=x^{-8}y^{7}\). Thus (c=1), (r=-8), (s=7), and (c+r+s=0).

Step 2

Why this answer is correct

The correct answer is B. (2). The expression is \(9x^{-4}y^{6}\cdot\frac{1}{9}x^{-4}y=x^{-8}y^{7}\). Thus (c=1), (r=-8), (s=7), and (c+r+s=0).

Step 3

Exam Tip

अभिव्यक्ति \(9x^{-4}y^{6}\cdot\frac{1}{9}x^{-4}y=x^{-8}y^{7}\) है। इसलिए (c=1), (r=-8), (s=7), और (c+r+s=0) होता है।

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(\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{-\frac{1}{3}}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{-\frac{1}{3}})?

Explanation opens after your attempt
Correct Answer

A. \(\frac{4x^{3}y^{4}}{5}\)

Step 1

Concept

We get (\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{\frac{1}{3}}=\frac{5x^{-3}}{4y^{4}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{4x^{3}y^{4}}{5}\).

Step 2

Why this answer is correct

The correct answer is A. \(\frac{4x^{3}y^{4}}{5}\). We get (\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{\frac{1}{3}}=\frac{5x^{-3}}{4y^{4}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{4x^{3}y^{4}}{5}\).

Step 3

Exam Tip

(\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{\frac{1}{3}}=\frac{5x^{-3}}{4y^{4}})। \(-\frac{1}{3}\) घात लेने पर व्युत्क्रम \(\frac{4x^{3}y^{4}}{5}\) मिलता है।

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(\left\(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}\right\)^{-1}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}\right\)^{-1})?

Explanation opens after your attempt
Correct Answer

A. \(9r^{6}s^{-8}\)

Step 1

Concept

Inside, \(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}=\frac{1}{9}r^{-6}s^{8}\). Raising to (-1) gives \(9r^{6}s^{-8}\).

Step 2

Why this answer is correct

The correct answer is A. \(9r^{6}s^{-8}\). Inside, \(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}=\frac{1}{9}r^{-6}s^{8}\). Raising to (-1) gives \(9r^{6}s^{-8}\).

Step 3

Exam Tip

अंदर \(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}=\frac{1}{9}r^{-6}s^{8}\) है। (-1) घात लेने पर \(9r^{6}s^{-8}\) मिलता है।

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\(\sqrt[3]{343a^{15}b^{12}}\) का सरल रूप क्या है?

What is the simplified form of \(\sqrt[3]{343a^{15}b^{12}}\)?

Explanation opens after your attempt
Correct Answer

A. \(7a^{5}b^{4}\)

Step 1

Concept

We have \(\sqrt[3]{343}=7\), \(\sqrt[3]{a^{15}}=a^{5}\), and \(\sqrt[3]{b^{12}}=b^{4}\). In exams, divide exponents by (3) under a cube root.

Step 2

Why this answer is correct

The correct answer is A. \(7a^{5}b^{4}\). We have \(\sqrt[3]{343}=7\), \(\sqrt[3]{a^{15}}=a^{5}\), and \(\sqrt[3]{b^{12}}=b^{4}\). In exams, divide exponents by (3) under a cube root.

Step 3

Exam Tip

\(\sqrt[3]{343}=7\), \(\sqrt[3]{a^{15}}=a^{5}\), और \(\sqrt[3]{b^{12}}=b^{4}\)। परीक्षा में घनमूल में घातों को (3) से भाग दें।

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यदि (\left\(x^{4}y^{-3}\right\)^{k}=x^{16}y^{-12}), तो (k) का मान क्या है?

If (\left\(x^{4}y^{-3}\right\)^{k}=x^{16}y^{-12}), what is the value of (k)?

Explanation opens after your attempt
Correct Answer

C. (4)

Step 1

Concept

The left side has exponents (4k) and (-3k). Both (4k=16) and (-3k=-12) give (k=4).

Step 2

Why this answer is correct

The correct answer is C. (4). The left side has exponents (4k) and (-3k). Both (4k=16) and (-3k=-12) give (k=4).

Step 3

Exam Tip

बाएँ पक्ष में घातें (4k) और (-3k) हैं। (4k=16) और (-3k=-12) दोनों से (k=4) मिलता है।

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(\left\(\frac{p^{-5}q^{4}}{p^{-1}q^{-2}}\right\)^{-2}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{p^{-5}q^{4}}{p^{-1}q^{-2}}\right\)^{-2})?

Explanation opens after your attempt
Correct Answer

A. \(p^{8}q^{-12}\)

Step 1

Concept

Inside, (p^{-5-(-1)}q^{4-(-2)}=p^{-4}q^{6}). Raising to (-2) gives \(p^{8}q^{-12}\).

Step 2

Why this answer is correct

The correct answer is A. \(p^{8}q^{-12}\). Inside, (p^{-5-(-1)}q^{4-(-2)}=p^{-4}q^{6}). Raising to (-2) gives \(p^{8}q^{-12}\).

Step 3

Exam Tip

अंदर (p^{-5-(-1)}q^{4-(-2)}=p^{-4}q^{6}) है। (-2) घात देने पर \(p^{8}q^{-12}\) मिलता है।

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यदि (\left\(2x^{-1}y^{2}\right\)^{3}\cdot\left\(4x^{2}y^{-1}\right\)^{-1}) को \(cx^{r}y^{s}\) लिखा जाए, तो (c+r+s) का मान क्या है?

If (\left\(2x^{-1}y^{2}\right\)^{3}\cdot\left\(4x^{2}y^{-1}\right\)^{-1}) is written as \(cx^{r}y^{s}\), what is the value of (c+r+s)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{17}{4}\)

Step 1

Concept

The expression is \(8x^{-3}y^{6}\cdot\frac{1}{4}x^{-2}y=2x^{-5}y^{7}\). Hence (c+r+s=2-5+7=4).

Step 2

Why this answer is correct

The correct answer is A. \(\frac{17}{4}\). The expression is \(8x^{-3}y^{6}\cdot\frac{1}{4}x^{-2}y=2x^{-5}y^{7}\). Hence (c+r+s=2-5+7=4).

Step 3

Exam Tip

अभिव्यक्ति \(8x^{-3}y^{6}\cdot\frac{1}{4}x^{-2}y=;2x^{-5}y^{7}\) है। इसलिए (c+r+s=2-5+7=4) है।

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(\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{-\frac{1}{3}}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{-\frac{1}{3}})?

Explanation opens after your attempt
Correct Answer

A. \(\frac{3x^{2}y^{3}}{4}\)

Step 1

Concept

We get (\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{\frac{1}{3}}=\frac{4x^{-2}}{3y^{3}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{3x^{2}y^{3}}{4}\).

Step 2

Why this answer is correct

The correct answer is A. \(\frac{3x^{2}y^{3}}{4}\). We get (\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{\frac{1}{3}}=\frac{4x^{-2}}{3y^{3}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{3x^{2}y^{3}}{4}\).

Step 3

Exam Tip

(\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{\frac{1}{3}}=\frac{4x^{-2}}{3y^{3}})। \(-\frac{1}{3}\) घात लेने पर व्युत्क्रम \(\frac{3x^{2}y^{3}}{4}\) मिलता है।

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(\left\(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}\right\)^{-1}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}\right\)^{-1})?

Explanation opens after your attempt
Correct Answer

A. \(7r^{5}s^{-6}\)

Step 1

Concept

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।

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\(\sqrt[3]{216a^{12}b^{9}}\) का सरल रूप क्या है?

What is the simplified form of \(\sqrt[3]{216a^{12}b^{9}}\)?

Explanation opens after your attempt
Correct Answer

A. \(6a^{4}b^{3}\)

Step 1

Concept

We have \(\sqrt[3]{216}=6\), \(\sqrt[3]{a^{12}}=a^{4}\), and \(\sqrt[3]{b^{9}}=b^{3}\). In exams, divide exponents by (3) under a cube root.

Step 2

Why this answer is correct

The correct answer is A. \(6a^{4}b^{3}\). We have \(\sqrt[3]{216}=6\), \(\sqrt[3]{a^{12}}=a^{4}\), and \(\sqrt[3]{b^{9}}=b^{3}\). In exams, divide exponents by (3) under a cube root.

Step 3

Exam Tip

\(\sqrt[3]{216}=6\), \(\sqrt[3]{a^{12}}=a^{4}\), और \(\sqrt[3]{b^{9}}=b^{3}\)। परीक्षा में घनमूल में घातों को (3) से भाग दें।

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यदि (\left\(x^{-3}y^{2}\right\)^{k}=x^{-12}y^{8}), तो (k) का मान क्या है?

If (\left\(x^{-3}y^{2}\right\)^{k}=x^{-12}y^{8}), what is the value of (k)?

Explanation opens after your attempt
Correct Answer

C. (4)

Step 1

Concept

The left side has exponents (-3k) and (2k). Both (-3k=-12) and (2k=8) give (k=4).

Step 2

Why this answer is correct

The correct answer is C. (4). The left side has exponents (-3k) and (2k). Both (-3k=-12) and (2k=8) give (k=4).

Step 3

Exam Tip

बाएँ पक्ष में घातें (-3k) और (2k) हैं। (-3k=-12) और (2k=8) दोनों से (k=4) मिलता है।

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(\left\(\frac{m^{-4}n^{3}}{m^{2}n^{-5}}\right\)^{-1}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{m^{-4}n^{3}}{m^{2}n^{-5}}\right\)^{-1})?

Explanation opens after your attempt
Correct Answer

B. \(m^{6}n^{-8}\)

Step 1

Concept

Inside, \(m^{-4-2}n^{3-(-5)}=m^{-6}n^{8}\). Raising to (-1) gives \(m^{6}n^{-8}\).

Step 2

Why this answer is correct

The correct answer is B. \(m^{6}n^{-8}\). Inside, \(m^{-4-2}n^{3-(-5)}=m^{-6}n^{8}\). Raising to (-1) gives \(m^{6}n^{-8}\).

Step 3

Exam Tip

अंदर \(m^{-4-2}n^{3-(-5)}=m^{-6}n^{8}\) है। (-1) घात लेने पर \(m^{6}n^{-8}\) मिलता है।

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(\left\(\frac{27x^{-3}}{8y^{6}}\right\)^{-\frac{1}{3}}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{27x^{-3}}{8y^{6}}\right\)^{-\frac{1}{3}})?

Explanation opens after your attempt
Correct Answer

A. \(\frac{2xy^{2}}{3}\)

Step 1

Concept

We get (\left\(\frac{27x^{-3}}{8y^{6}}\right\)^{\frac{1}{3}}=\frac{3x^{-1}}{2y^{2}}), so the power \(-\frac{1}{3}\) gives its reciprocal \(\frac{2xy^{2}}{3}\). In exams, treat the negative fractional power as a reciprocal after rooting.

Step 2

Why this answer is correct

The correct answer is A. \(\frac{2xy^{2}}{3}\). We get (\left\(\frac{27x^{-3}}{8y^{6}}\right\)^{\frac{1}{3}}=\frac{3x^{-1}}{2y^{2}}), so the power \(-\frac{1}{3}\) gives its reciprocal \(\frac{2xy^{2}}{3}\). In exams, treat the negative fractional power as a reciprocal after rooting.

Step 3

Exam Tip

(\left\(\frac{27x^{-3}}{8y^{6}}\right\)^{\frac{1}{3}}=\frac{3x^{-1}}{2y^{2}}), इसलिए \(-\frac{1}{3}\) घात देने पर उसका व्युत्क्रम \(\frac{2xy^{2}}{3}\) है। परीक्षा में भिन्न घात के बाद ऋणात्मक संकेत को व्युत्क्रम मानें।

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(\left\(\frac{5m^{-2}n^{3}}{25m^{4}n^{-1}}\right\)^{-1}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{5m^{-2}n^{3}}{25m^{4}n^{-1}}\right\)^{-1})?

Explanation opens after your attempt
Correct Answer

A. \(5m^{6}n^{-4}\)

Step 1

Concept

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।

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\(\sqrt[3]{125a^{9}b^{6}}\) का सरल रूप क्या है?

What is the simplified form of \(\sqrt[3]{125a^{9}b^{6}}\)?

Explanation opens after your attempt
Correct Answer

A. \(5a^{3}b^{2}\)

Step 1

Concept

We have \(\sqrt[3]{125}=5\), \(\sqrt[3]{a^{9}}=a^{3}\), and \(\sqrt[3]{b^{6}}=b^{2}\). In exams, divide exponents by (3) under a cube root.

Step 2

Why this answer is correct

The correct answer is A. \(5a^{3}b^{2}\). We have \(\sqrt[3]{125}=5\), \(\sqrt[3]{a^{9}}=a^{3}\), and \(\sqrt[3]{b^{6}}=b^{2}\). In exams, divide exponents by (3) under a cube root.

Step 3

Exam Tip

\(\sqrt[3]{125}=5\), \(\sqrt[3]{a^{9}}=a^{3}\), और \(\sqrt[3]{b^{6}}=b^{2}\)। परीक्षा में घनमूल में घातों को (3) से भाग दें।

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यदि (\left\(x^{2}y^{-1}\right\)^{k}=x^{10}y^{-5}), तो (k) का मान क्या है?

If (\left\(x^{2}y^{-1}\right\)^{k}=x^{10}y^{-5}), what is the value of (k)?

Explanation opens after your attempt
Correct Answer

C. (5)

Step 1

Concept

The left side has exponents (2k) and (-k). Both (2k=10) and (-k=-5) give (k=5).

Step 2

Why this answer is correct

The correct answer is C. (5). The left side has exponents (2k) and (-k). Both (2k=10) and (-k=-5) give (k=5).

Step 3

Exam Tip

बाएँ पक्ष में घातें (2k) और (-k) हैं। (2k=10) और (-k=-5) दोनों से (k=5) मिलता है।

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(\left\(\frac{4x^{3}y^{-2}}{2x^{-1}y^{4}}\right\)^{2}\cdot\frac{y^{12}}{x^{4}}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{4x^{3}y^{-2}}{2x^{-1}y^{4}}\right\)^{2}\cdot\frac{y^{12}}{x^{4}})?

Explanation opens after your attempt
Correct Answer

A. \(4x^{4}\)

Step 1

Concept

Inside, \(\frac{4x^{3}y^{-2}}{2x^{-1}y^{4}}=2x^{4}y^{-6}\), and its square is \(4x^{8}y^{-12}\). Multiplying by \(\frac{y^{12}}{x^{4}}\) gives \(4x^{4}\).

Step 2

Why this answer is correct

The correct answer is A. \(4x^{4}\). Inside, \(\frac{4x^{3}y^{-2}}{2x^{-1}y^{4}}=2x^{4}y^{-6}\), and its square is \(4x^{8}y^{-12}\). Multiplying by \(\frac{y^{12}}{x^{4}}\) gives \(4x^{4}\).

Step 3

Exam Tip

अंदर \(\frac{4x^{3}y^{-2}}{2x^{-1}y^{4}}=2x^{4}y^{-6}\), इसका वर्ग \(4x^{8}y^{-12}\) है। फिर \(\frac{y^{12}}{x^{4}}\) से गुणा करने पर \(4x^{4}\) मिलता है।

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(\left\(\frac{a^{-3}b^{2}}{a^{2}b^{-4}}\right\)^{-2}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{a^{-3}b^{2}}{a^{2}b^{-4}}\right\)^{-2})?

Explanation opens after your attempt
Correct Answer

A. \(a^{10}b^{-12}\)

Step 1

Concept

Inside, \(a^{-3-2}b^{2-(-4)}=a^{-5}b^{6}\), so raising to (-2) gives \(a^{10}b^{-12}\). In exams, a negative outer power changes the signs of both exponents.

Step 2

Why this answer is correct

The correct answer is A. \(a^{10}b^{-12}\). Inside, \(a^{-3-2}b^{2-(-4)}=a^{-5}b^{6}\), so raising to (-2) gives \(a^{10}b^{-12}\). In exams, a negative outer power changes the signs of both exponents.

Step 3

Exam Tip

अंदर \(a^{-3-2}b^{2-(-4)}=a^{-5}b^{6}\) है, इसलिए (-2) घात देने पर \(a^{10}b^{-12}\) मिलता है। परीक्षा में ऋणात्मक घात पर दोनों घातों के चिह्न बदलते हैं।

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(\left\(\frac{3x^{-2}}{y^{-1}}\right\)^{3}\cdot\frac{y^{2}}{27}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{3x^{-2}}{y^{-1}}\right\)^{3}\cdot\frac{y^{2}}{27})?

Explanation opens after your attempt
Correct Answer

A. \(x^{-6}y^{5}\)

Step 1

Concept

\(\frac{3x^{-2}}{y^{-1}}=3x^{-2}y\), its cube is \(27x^{-6}y^{3}\), and multiplying by \(\frac{y^{2}}{27}\) gives \(x^{-6}y^{5}\). In exams, turn division by a negative power into multiplication.

Step 2

Why this answer is correct

The correct answer is A. \(x^{-6}y^{5}\). \(\frac{3x^{-2}}{y^{-1}}=3x^{-2}y\), its cube is \(27x^{-6}y^{3}\), and multiplying by \(\frac{y^{2}}{27}\) gives \(x^{-6}y^{5}\). In exams, turn division by a negative power into multiplication.

Step 3

Exam Tip

\(\frac{3x^{-2}}{y^{-1}}=3x^{-2}y\), इसका घन \(27x^{-6}y^{3}\) है, फिर \(\frac{y^{2}}{27}\) से गुणा करने पर \(x^{-6}y^{5}\) मिलता है। परीक्षा में भाग को ऋणात्मक घात से गुणा में बदलें।

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यदि (\left\(x^{2}y^{3}\right\)^{n}=x^{8}y^{12}), तो (n) का मान क्या है?

If (\left\(x^{2}y^{3}\right\)^{n}=x^{8}y^{12}), what is the value of (n)?

Explanation opens after your attempt
Correct Answer

C. (4)

Step 1

Concept

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।

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(\left\(\frac{a^{3}b^{-2}}{a^{-1}b^{2}}\right\)^{2}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{a^{3}b^{-2}}{a^{-1}b^{2}}\right\)^{2})?

Explanation opens after your attempt
Correct Answer

A. \(a^{8}b^{-8}\)

Step 1

Concept

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।

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किस विकल्प में (\left\(ab^{-2}\right\)^{3}\cdot a^{-1}b^{5}) का सही सरल रूप है?

Which option gives the correct simplified form of (\left\(ab^{-2}\right\)^{3}\cdot a^{-1}b^{5})?

Explanation opens after your attempt
Correct Answer

A. \(a^{2}b^{-1}\)

Step 1

Concept

We have (\left\(ab^{-2}\right\)^{3}=a^{3}b^{-6}), and multiplying by \(a^{-1}b^{5}\) gives \(a^{2}b^{-1}\). In exams, add exponents separately for each variable.

Step 2

Why this answer is correct

The correct answer is A. \(a^{2}b^{-1}\). We have (\left\(ab^{-2}\right\)^{3}=a^{3}b^{-6}), and multiplying by \(a^{-1}b^{5}\) gives \(a^{2}b^{-1}\). In exams, add exponents separately for each variable.

Step 3

Exam Tip

(\left\(ab^{-2}\right\)^{3}=a^{3}b^{-6}), फिर \(a^{-1}b^{5}\) से गुणा करने पर \(a^{2}b^{-1}\) मिलता है। परीक्षा में हर चर की घात अलग-अलग जोड़ें।

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(\left\(\frac{4x^{2}y^{-3}}{2x^{-1}y}\right\)^{-2}) का सरल रूप क्या है, जहाँ \(x\neq0\) और \(y\neq0\)?

What is the simplified form of (\left\(\frac{4x^{2}y^{-3}}{2x^{-1}y}\right\)^{-2}), where \(x\neq0\) and \(y\neq0\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{y^{8}}{4x^{6}}\)

Step 1

Concept

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।

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यदि \(x \neq 0\), तो (\dfrac{(2x)3\(3x^{-2}\)}{12x^{-1}}) का सरल रूप क्या है?

If \(x \neq 0\), what is the simplified form of (\dfrac{(2x)3\(3x^{-2}\)}{12x^{-1}})?

Explanation opens after your attempt
Correct Answer

A. \(,2x^2,\)

Step 1

Concept

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।

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(\(-4a^2b^{-3}\)\(3a^{-1}b^5\)) का गुणनफल क्या है?

What is the product of (\(-4a^2b^{-3}\)\(3a^{-1}b^5\))?

Explanation opens after your attempt
Correct Answer

A. \(,-12ab^2,\)

Step 1

Concept

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।

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यदि \(a \neq 0\) और \(b \neq 0\), तो \(\dfrac{6a^3b^2}{2ab^{-1}}\) का सरल रूप क्या है?

If \(a \neq 0\) and \(b \neq 0\), what is the simplified form of \(\dfrac{6a^3b^2}{2ab^{-1}}\)?

Explanation opens after your attempt
Correct Answer

A. \(,3a^2b^3,\)

Step 1

Concept

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।

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(\(2x^2y\)\(-3xy^2\)) का गुणनफल क्या है?

What is the product of (\(2x^2y\)\(-3xy^2\))?

Explanation opens after your attempt
Correct Answer

A. \(,-6x^3y^3,\)

Step 1

Concept

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।

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