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The square root of$$ \frac{(0.75)^{3}}{1-0.75}+[0.75+(0.75)^{2}+1]$$
let us consider 0.75 as x. then equation becomes
$$ \sqrt[2]{x^3/(1-x) +(x+x^2+1)}$$
= $$ \sqrt[2]{[x^3 +(1-x)(x+x^2+1)] \div (1-x)}$$
= $$ \sqrt[2]{[x^3 +(1-x^3)] \div (1-x)}$$
= $$ \sqrt[2]{1 \div (1-x)}$$
If x is replaced by 0.75, we get
= $$ \sqrt[2]{1 \div 0.25}$$
= $$ \sqrt[2]4$$
=2
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