The square root of frac{(0.75)^3}{1-(0.75)}+l | vedclass

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Question

(b)

Let $x=0.75$

According to the question, $\frac{x^3}{1-x}+\left(x+x^2+1\right)$

$=\frac{x^3+(1-x)\left(1+x+x^2\right)}{1-x}$

$=\frac{x^

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(b)

Let $x=0.75$

According to the question, $\frac{x^3}{1-x}+\left(x+x^2+1\right)$

$=\frac{x^3+(1-x)\left(1+x+x^2\right)}{1-x}$

$=\frac{x^3+1-x^3}{1-x}=\frac{1}{1-x}$

Now, put the value of $x$

$\frac{1}{1-0.75} =\frac{1}{0.25}$

$=\frac{100}{25}=4$

So, square root of the equation $=\sqrt{4}=2$

The square root of $\frac{(0.75)^3}{1-(0.75)}+\left[0.75+(0.75)^2+1\right]$ is

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