Giải giúp tôi bai toán này

Giải thích các bước giải:

a.Ta có:

$A=(\dfrac{\sqrt{x}+1}{x-\sqrt{x}+1}-\dfrac{x-3\sqrt{x}}{x\sqrt{x}+1}):\dfrac{5\sqrt{x}+1}{3x-3\sqrt{x}+3}$

$\to A=\dfrac{(\sqrt{x}+1)^2-(x-3\sqrt{x})}{(\sqrt{x}+1)(x-\sqrt{x}+1)}\cdot \dfrac{3(x-\sqrt{x}+1)}{5\sqrt{x}+1}$

$\to A=\dfrac{x+2\sqrt{x}+1-x+3\sqrt{x}}{(\sqrt{x}+1)(x-\sqrt{x}+1)}\cdot \dfrac{3(x-\sqrt{x}+1)}{5\sqrt{x}+1}$

$\to A=\dfrac{5\sqrt{x}+1}{(\sqrt{x}+1)(x-\sqrt{x}+1)}\cdot \dfrac{3(x-\sqrt{x}+1)}{5\sqrt{x}+1}$

$\to A=\dfrac{3}{\sqrt{x}+1}$

b.Ta có:

$\begin{cases}x_1+x_2=12\\x_1x_2=9\end{cases}$

$x_2^2-12x_2+9=0$

$\to x_2^2=12x_2-9$

$\to x_2^3=12x_2^2-9x_2$

$\to x_2^3=12(12x_2-9)-9x_2$

$\to x_2^3=135x_2-108$

$\to x_2^3-134x_2+108=x_2$

$\to A=\sqrt{x_1}+\sqrt{x_2}$

$\to A^2=(\sqrt{x_1}+\sqrt{x_2})^2$

$\to A^2=x_1+x_2+2\sqrt{x_1x_2}$

$\to A^2=12+2\sqrt9$

$\to A^2=18$

$\to A=3\sqrt2$