Cho hàm số y = f(x) = ( căn bậc hai x + 1) / ( căn bậc hai x - 1)

* Hướng dẫn giải

❶ Hàm số xác định khi

$\left\{\begin{array}{l}x\ge 0\\ \sqrt{x}-1\ne 0\end{array}\right.\iff \left\{\begin{array}{l}x\ge 0\\ \sqrt{x}\ne 1\end{array}\right.\iff \left\{\begin{array}{l}x\ge 0\\ x\ne 1\end{array}\right.\iff 0\le x\ne 1$

Vậy điều kiện xác định của hàm số là: $D=\left(0;+\infty \right)\backslash \left\{1\right\}$

❷ 

$\begin{array}{l}f(4+2\sqrt{3})=\frac{\sqrt{4+2\sqrt{3}}+1}{\sqrt{4+2\sqrt{3}}-1}\\ =\frac{\sqrt{{\left(\sqrt{3}+1\right)}^2}+1}{\sqrt{{\left(\sqrt{3}+1\right)}^2}-1}\\ =\frac{\sqrt{3}+1+1}{\sqrt{3}+1-1}\\ =\frac{\sqrt{3}+2}{\sqrt{3}}\end{array}$

❸ Để $f\left(x\right)=\sqrt{3}$ thì:

$\begin{array}{l}f\left(x\right)=\sqrt{3}\iff \frac{\sqrt{x}+1}{\sqrt{x}-1}=\sqrt{3}\\ \iff \sqrt{x}+1=\sqrt{3}.\sqrt{x}-\sqrt{3}\\ \iff \left(\sqrt{3}-1\right)\sqrt{x}=\sqrt{3}+1\\ \iff \sqrt{x}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\\ \iff x=\frac{4+2\sqrt{3}}{4-2\sqrt{3}}\end{array}$

❹ Để $f\left(x\right)=f\left(x^2\right)$ thì:

$\begin{array}{l}f\left(x\right)=f\left(x^2\right)\iff \frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{x-1}{x+1}\\ \iff \frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x-1}\\ \iff x+1={\left(\sqrt{x}-1\right)}^2\\ \iff \sqrt{x}=0\\ \iff x=0\end{array}$