mn giải giúp mình bài này nhé mình cảm ơn

1. 
$\displaystyle B=\frac{6+2\sqrt{3}}{\sqrt{3} +1} -\frac{2}{\sqrt{5} -\sqrt{3}} +\sqrt{8-2\sqrt{15}}$
$\displaystyle B=\frac{\left( 6+2\sqrt{3}\right)\left(\sqrt{3} -1\right)}{\left(\sqrt{3} +1\right)\left(\sqrt{3} -1\right)} -\frac{2\left(\sqrt{5} +\sqrt{3}\right)}{\left(\sqrt{5} -\sqrt{3}\right)\left(\sqrt{5} +\sqrt{3}\right)} +\sqrt{5-2\sqrt{5} \cdot \sqrt{3} +3}$
$\displaystyle B=\frac{6\sqrt{3} -6+6-2\sqrt{3}}{\left(\sqrt{3}\right)^{2} -1} -\frac{2\left(\sqrt{5} +\sqrt{3}\right)}{\left(\sqrt{5}\right)^{2} -\left(\sqrt{3}\right)^{2}} +\sqrt{\left(\sqrt{5} -\sqrt{3}\right)^{2}}$
$\displaystyle B=\frac{4\sqrt{3}}{3-1} -\frac{2\left(\sqrt{5} +\sqrt{3}\right)}{5-3} +\left(\sqrt{5} -\sqrt{3}\right)$
$\displaystyle B=\frac{4\sqrt{3}}{2} -\frac{2\left(\sqrt{5} +\sqrt{3}\right)}{2} +\sqrt{5} -\sqrt{3}$
$\displaystyle B=2\sqrt{3} -\left(\sqrt{5} +\sqrt{3}\right) +\sqrt{5} -\sqrt{3}$
$\displaystyle B=2\sqrt{3} -\sqrt{5} -\sqrt{3} +\sqrt{5} -\sqrt{3}$
$\displaystyle B=0$
2. 
ĐKXĐ $\displaystyle x\geqslant 0,\ x\neq 4$
a. Ta có
$\displaystyle A=\frac{\sqrt{x} +1}{\sqrt{x} -2} +\frac{2\sqrt{x}}{\sqrt{x} +2} +\frac{2+5\sqrt{x}}{4-x}$
$\displaystyle A=\frac{\sqrt{x} +1}{\sqrt{x} -2} +\frac{2\sqrt{x}}{\sqrt{x} +2} +\frac{2+5\sqrt{x}}{-( x-4)}$
$\displaystyle A=\frac{\left(\sqrt{x} +1\right)\left(\sqrt{x} +2\right)}{\left(\sqrt{x} -2\right)\left(\sqrt{x} +2\right)} +\frac{2\sqrt{x}\left(\sqrt{x} -2\right)}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)} -\frac{2+5\sqrt{x}}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)}$
$\displaystyle A=\frac{x+3\sqrt{x} +2}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)} +\frac{2x-4\sqrt{x}}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)} -\frac{2+5\sqrt{x}}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)}$
$\displaystyle A=\frac{\left( x+3\sqrt{x} +2\right) +\left( 2x-4\sqrt{x}\right) -\left( 2+5\sqrt{x}\right)}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)}$
$\displaystyle A=\frac{x+3\sqrt{x} +2+2x-4\sqrt{x} -2-5\sqrt{x}}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)}$
$\displaystyle A=\frac{3x-6\sqrt{x}}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)}$
$\displaystyle A=\frac{3\sqrt{x}\left(\sqrt{x} -2\right)}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)}$
$\displaystyle A=\frac{3\sqrt{x}}{\sqrt{x} +2}$
Vậy $\displaystyle A=\frac{3\sqrt{x}}{\sqrt{x} +2}$ với $\displaystyle x\geqslant 0,\ x\neq 4$
b. Để $\displaystyle A=2$ thì $\displaystyle \frac{3\sqrt{x}}{\sqrt{x} +2} =2$
Ta có $\displaystyle \frac{3\sqrt{x}}{\sqrt{x} +2} =2\Rightarrow 3\sqrt{x} =2\left(\sqrt{x} +2\right) \Rightarrow 3\sqrt{x} =2\sqrt{x} +4$
$\displaystyle \Rightarrow \sqrt{x} =4\Rightarrow \left(\sqrt{x}\right)^{2} =4^{2} \Rightarrow x=16$ (thỏa mãn đkxđ)
Vậy $\displaystyle x=16$ thì $\displaystyle A=2$