MCQ
The square root of $\frac{(0.75)^3}{1-(0.75)}+\left[0.75+(0.75)^2+1\right]$ is
- A$1$
- ✓$2$
- C$3$
- D$4$
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 $=\sqrt{4}=2$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.