Cho biểu thức Q = $\left(\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{\sqrt{x}}{1+\sqrt{x}}\right)+\frac{3-\sqrt{x}}{x-1}$  với x ≥ 0 và x ≠ 1.

a) Rút gọn Q.

b) Tìm x để Q = – 1.

a) Q =  $\left(\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{\sqrt{x}}{1+\sqrt{x}}\right)+\frac{3-\sqrt{x}}{x-1}$

Q =   $\frac{-\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{1+\sqrt{x}}+\frac{3-\sqrt{x}}{\left(1+\sqrt{x}\right)\left(\sqrt{x}-1\right)}$

Q = $\frac{-\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}\left(\sqrt{x}-1\right)+3-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}$

Q = $\frac{-x-\sqrt{x}+x-\sqrt{x}+3-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}$

Q = $\frac{3\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}$

Q = $\frac{-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=-\frac{3}{\sqrt{x}+1}$

b) Để Q = – 1 thì $-\frac{3}{\sqrt{x}+1}=-1$

⇔ $\sqrt{x}+1=3$

⇔ $\sqrt{x}=2$

⇔ x = 4.