The correct option is B 9b^616a^4 [4a^23b^3]^ 2 =[3b^34a^2]^2 #160; #160; #160; #160; #160; #160; #160; #160; #160; ...

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Simplify [4...

Question

Simplify $[\frac{4 a^{2}}{3 b^{3}}]^{- 2}$

\frac{16 a^{4}}{9 b^{6}}

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\frac{9 b^{6}}{16 a^{4}}

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\frac{16 b^{4}}{9 a^{6}}

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\frac{16 b^{6}}{9 a^{4}}

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Solution

The correct option is B $\frac{9 b^{6}}{16 a^{4}}$
${[\frac{4 a^{2}}{3 b^{3}}]}^{- 2}$ $= {[\frac{3 b^{3}}{4 a^{2}}]}^{2}$

${[\frac{4 a^{2}}{3 b^{3}}]}^{- 2}$ $= [\frac{(3 b^{3})^{2}}{(4 a^{2})^{2}}]$

${[\frac{4 a^{2}}{3 b^{3}}]}^{- 2}$ $= [\frac{9 b^{6}}{16 a^{4}}]$

$O p t : [B]$

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Similar questions

If $2 a + \frac{1}{2 a} = 8$ , find :

(i) $4 a^{2} + \frac{1}{4 a^{2}}$

(ii) $16 a^{4} + \frac{1}{16 a^{4}}$

If $2 a + \frac{1}{2 a} = 3$ , then $16 a^{4} + \frac{1}{16 a^{4}} = ?$

2 a + \frac{1}{2 a} = 8

16 a^{4} + \frac{1}{16 a^{4}}

2 a - \frac{1}{2 a} = 3

16 a^{4} + \frac{1}{16 a^{4}}

16 a^{4} - 9 b^{4}

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