Giải thích các bước giải:
a.Ta có:
$B=\dfrac{\sqrt{x}\left(1-x\right)^2}{1+x}:\left(\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\right)\cdot \left(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}-\sqrt{x}\right)\right)$
$\to B=\dfrac{\sqrt{x}\left(1-x\right)^2}{1+x}:\left(\left(\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}-1}+\sqrt{x}\right)\cdot \left(\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}-\sqrt{x}\right)\right)$
$\to B=\dfrac{\sqrt{x}\left(1-x\right)^2}{1+x}:\left(\left(x+\sqrt{x}+1+\sqrt{x}\right)\cdot \left(x-\sqrt{x}+1-\sqrt{x}\right)\right)$
$\to B=\dfrac{\sqrt{x}\left(1-x\right)^2}{1+x}:\left(\left(x+2\sqrt{x}+1\right)\cdot \left(x-2\sqrt{x}+1\right)\right)$
$\to B=\dfrac{\sqrt{x}\left(1-x\right)^2}{1+x}:\left(\left(\sqrt{x}+1\right)^2\cdot \left(\sqrt{x}-1\right)^2\right)$
$\to B=\dfrac{\sqrt{x}\left(x-1\right)^2}{1+x}:\left(\left(\sqrt{x}+1\right)\cdot \left(\sqrt{x}-1\right)\right)^2$
$\to B=\dfrac{\sqrt{x}\left(x-1\right)^2}{1+x}:\left(x-1\right)^2$
$\to B=\dfrac{\sqrt{x}\left(x-1\right)^2}{1+x}\cdot\dfrac{1}{\left(x-1\right)^2}$
$\to B=\dfrac{\sqrt{x}}{1+x}$
b.Để $B=\dfrac25$
$\to \dfrac{\sqrt{x}}{1+x}=\dfrac25$
$\to 5\sqrt{x}=2\left(1+x\right)$
$\to 2x-5\sqrt{x}+2=0$
$\to \left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0$
$\to \sqrt{x}\in\{2,\dfrac12\}$
$\to x\in\{4,\dfrac14\}$
