Giải bài 7 trang 62 – SGK môn Đại số lớp 10

a) Điều kiện xác định: $x\ge \dfrac{-6}{5}$
Ta có:
$\begin{aligned} & \sqrt{5x+6}=x-6 \\ & \Leftrightarrow \left\{ \begin{aligned} & x-6\ge 0 \\ & 5x+6={{\left( x-6 \right)}^{2}} \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & x\ge 6 \\ & {{x}^{2}}-12x+36-5x-6=0 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & x\ge 6 \\ & {{x}^{2}}-17x+30=0 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & x\ge 6 \\ & \left[ \begin{aligned} & x=15 \\ & x=2 \\ \end{aligned} \right. \\ \end{aligned} \right. \\ & \Leftrightarrow x=15 \\ \end{aligned}$
Vậy $S=\left\{ 15 \right\}$
b) ĐKXĐ: $\left\{ \begin{aligned} & 3-x\ge 0 \\ & x+2\ge 0 \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & x\le 3 \\ & x\ge -2 \\ \end{aligned} \right.\Leftrightarrow -2\le x\le 3$
Ta có:
$\begin{aligned} & \sqrt{3-x}=\sqrt{x+2}+1 \\ & \Leftrightarrow 3-x=x+2+2\sqrt{x+2}+1 \\ & \Leftrightarrow 2\sqrt{x+2}=-2x \\ & \Leftrightarrow \left\{ \begin{aligned} & x\le 0 \\ & x+2={{x}^{2}} \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & x\le 0 \\ & {{x}^{2}}-x-2=0 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & x\le 0 \\ & \left[ \begin{aligned} & x=-1 \\ & x=2 \\ \end{aligned} \right. \\ \end{aligned} \right. \\ & \Leftrightarrow x=-1\,\,\left( \text{thỏa mãn} \right) \\ \end{aligned}$
Vậy $S=\left\{ -1 \right\}$
c) TXĐ: $D=\mathbb{R}$
Ta có:
$\begin{aligned} & \sqrt{2{{x}^{2}}+5}=x+2 \\ & \Leftrightarrow \left\{ \begin{aligned} & x+2\ge 0 \\ & 2{{x}^{2}}+5={{x}^{2}}+4x+4 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & x\ge -2 \\ & {{x}^{2}}-4x+1=0 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & x\ge -2 \\ & \left[ \begin{aligned} & x=2+\sqrt{3} \\ & x=2-\sqrt{3} \\ \end{aligned} \right. \\ \end{aligned} \right. \\ \end{aligned}$ 
Vậy $S=\left\{ 2-\sqrt{3};2+\sqrt{3} \right\}$
d) TXĐ: $D=\mathbb R$
$\begin{aligned} & \sqrt{4{{x}^{2}}+2x+10}=3x+1 \\ & \Leftrightarrow \left\{ \begin{aligned} & 3x+1\ge 0 \\ & 4{{x}^{2}}+2x+10={{\left( 3x+1 \right)}^{2}} \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & x\ge -\dfrac{1}{3} \\ & 4{{x}^{2}}+2x+10=9{{x}^{2}}+6x+1 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & x\ge -\dfrac{1}{3} \\ & 5{{x}^{2}}+4x-9=0 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & x\ge -\dfrac{1}{3} \\ & \left[ \begin{aligned} & x=1 \\ & x=-\dfrac{9}{5} \\ \end{aligned} \right. \\ \end{aligned} \right. \\ & \Leftrightarrow x=1 \\ \end{aligned}$
Vậy $S=\left\{ 1 \right\}$

Ghi nhớ:

$\sqrt{f(x)}=g(x)\\ \Leftrightarrow \left\{\begin{align} &g(x)\ge 0\\&f(x)=g^2(x)\\ \end{align}\right.$