Cho biểu thức A=[(xcănx+1)/(cănx-1)(căn x+1)-(x-1)/(căn x-1]:(căn x+ căn x/ cănx -1)

* Hướng dẫn giải

$A=\left[\frac{x\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{x-1}{\sqrt{x}-1}\right]:\left(\sqrt{x}+\frac{\sqrt{x}}{\sqrt{x}-1}\right)$ $=\left[\frac{\sqrt{x^3}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\right]:\frac{x-\sqrt{x}+\sqrt{x}}{\sqrt{x}-1}$ $=\left[\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\sqrt{x}-1\right].\frac{\sqrt{x}-1}{x}$$=\left(\frac{x-\sqrt{x}+1}{\sqrt{x}-1}+\frac{x-1}{\sqrt{x}-1}\right).\frac{\sqrt{x}-1}{x}$$=\frac{x-\sqrt{x}+1+x-1}{\sqrt{x}-1}.\frac{\sqrt{x}-1}{x}$$=\frac{x-\sqrt{x}+1+x-1}{\sqrt{x}-1}.\frac{\sqrt{x}-1}{x}$$=\frac{2x-\sqrt{x}}{x}=\frac{2\sqrt{x}-1}{\sqrt{x}}$

b. $A=3\iff \frac{2\sqrt{x}-1}{\sqrt{x}}=3\iff \sqrt{x}-1=3\sqrt{x}\iff \sqrt{x}=-1$