Question

pleasee anwere friends ​

Answer 1

Answer:

1/36

Step-by-step explanation:

this is the answer or

Answer 2

Given :

$\displaystyle{\sf \Bigg[\Big(\frac{-1}{3}\Big)^{-2}\Bigg]^{-2} \times \Bigg[\Big(\frac{2}{3}\Big)^{2}\Bigg]^{-2} \div \Bigg[\Big(\frac{3}{2}\Big)^{-1}\Bigg]^{-2}}$

Explanation :

A positive exponent tells us how many times to multiply a base number, and a negative exponent tells us how many times to divide a base number. We can rewrite negative exponents like x⁻ⁿ as $\sf \frac{1}{x^n}$ For example,

$\sf{2^{-4}= \frac{1}{2^4}=\frac{1}{16}}$

Solution :

$\rightarrow {\displaystyle{\sf \Bigg[\Big(\frac{-1}{3}\Big)^{-2}\Bigg]^{-2} \times \Bigg[\Big(\frac{2}{3}\Big)^{2}\Bigg]^{-2} \div \Bigg[\Big(\frac{3}{2}\Big)^{-1}\Bigg]^{-2}}}$

$\rightarrow \displaystyle{\sf \dfrac{\Big[\frac{-1}{3}\Big]^{-2 \times-2} \times \Big[\frac{2}{3}\Big]^{2 \times-2}}{\Big[\frac{3}{2}\Big]^{-1 \times -2}}}$

$\rightarrow \displaystyle{\sf \frac{\Big[\frac{-1}{3}\Big]^{4} \times \Big[\frac{2}{3}\Big]^{-4}}{\Big[\frac{3}{2}\Big]^{2}}}$

$\rightarrow \displaystyle{\sf \frac{\Big(\frac{(-1)^4}{(3)^4}\Big) \times \Big(\frac{(3)^4}{(2)^4}\Big)}{ \Big(\frac{(3)^2}{(2)^2}\Big)}}$

$\rightarrow \displaystyle{\sf \frac{\Big(\frac{1}{81}\Big) \times \Big(\frac{81}{16}\Big)}{\Big(\frac{9}{4}\Big)}}\implies \sf \displaystyle{\sf \frac{\Big(\frac{1}{16}\Big)}{\Big(\frac{4}{9}\Big)} }$

$\displaystyle {\rightarrow \Big(\frac{1}{16}\Big) \times \Big(\frac{4}{9}\Big)\implies \sf \frac{1}{36}}$

Required Answer :

Simplified form of $\displaystyle{\sf \Bigg[\Big(\frac{-1}{3}\Big)^{-2}\Bigg]^{-2} \times \Bigg[\Big(\frac{2}{3}\Big)^{2}\Bigg]^{-2} \div \Bigg[\Big(\frac{3}{2}\Big)^{-1}\Bigg]^{-2}}$ is $\displaystyle {\sf \frac{1}{36}}$