Giải thích các bước giải:
a.Ta có:
$C=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left(\left(\dfrac{1-x^3}{1-x}+x\right)\left(\dfrac{1+x^3}{1+x}\right)-x\right)$
$\to C=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left(\left(\dfrac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}+x\right)\left(\dfrac{\left(1+x\right)\left(1-x+x^2\right)}{1+x}\right)-x\right)$
$\to C=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left(\left(1+x+x^2+x\right)\left(1-x+x^2-x\right)\right)$
$\to C=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left(\left(1+2x+x^2\right)\left(1-2x+x^2\right)\right)$
$\to C=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left(\left(1+x\right)^2\left(1-x\right)^2\right)$
$\to C=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left(\left(1+x\right)\left(1-x\right)\right)^2$
$\to C=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left(1-x^2\right)^2$
$\to C=\dfrac{x}{1+x^2}$
b.Ta có:
$x=\sqrt{3+2\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1$
$\to x^2=3+2\sqrt{2}$
$\to C=\dfrac{\sqrt{2}+1}{1+3+2\sqrt{2}}$
$\to C=\dfrac{\sqrt{2}+1}{4+2\sqrt{2}}$
