Cho biểu thức: A = 1 : (x+2 / x căn bậc hai của x -1 + căn bậc hai của x

* Hướng dẫn giải

1. Điều kiện để biểu thức A có nghĩa: $\left\{\begin{array}{l}x\ge 0\\ x\sqrt{x}-1\ne 0\\ x-1\ne 0\end{array}\right.\iff \left\{\begin{array}{l}x\ge 0\\ x\ne 1\end{array}\right.$

2. Ta có:

$\begin{array}{l}A=1:\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\right)\\ =1:\left[\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}-1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\\ =1:\left[\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}-1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right]\\ =1:\left[\frac{x+2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(x+\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}-1\right)}\right]\\ =1:\left[\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}-1\right)}\right]\\ =1:\left[\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}-1\right)}\right]\\ =1:\left[\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}-1\right)}\right]\\ =\frac{x+\sqrt{x}+1}{\sqrt{x}}\end{array}$

3. Xét hiệu

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

Nhận thấy: $\left\{\begin{array}{l}{\left(\sqrt{x}-1\right)}^2\ge 0\\ \sqrt{x}\ge 0\\ 0<x\ne 1\end{array}\right.\iff \left\{\begin{array}{l}{\left(\sqrt{x}-1\right)}^2\ge 0\\ \sqrt{x}\ge 0\end{array}\right.\iff \frac{{\left(\sqrt{x}-1\right)}^2}{\sqrt{x}}\ge 0$

Do đó: $A-3\ge 0\iff A\ge 3$