Phương trình 

A

- Hướng dẫn giải

$\sqrt{2x^2+6x+1}=x+2$

$\iff \left\{\begin{array}{l}x+2\ge 0\\ 2x^2+6x+1={(x+2)}^2\end{array}\right.$

$$\begin{gathered} \iff \left\{\begin{array}{l}x\ge -2\\ x^2+2x-3=0\end{array}\right. \\ \iff \left\{\begin{array}{l}x\ge -2\\ x^2-x+3x-3=0\end{array}\right. \end{gathered}$$

$$\begin{gathered} \iff \left\{\begin{array}{l}x\ge -2\\ x(x-1)+3(x-1)=0\end{array}\right. \\ \iff \left\{\begin{array}{l}x\ge -2\\ (x-1)(x+3)=0\end{array}\right. \end{gathered}$$

$\iff \left\{\begin{array}{l}x\ge -2\\ \left[\begin{array}{c}x=1\\ x=-3\end{array}\right.\end{array}\iff x = 1\right.$

Vậy phương trình có nghiệm là x = 1.

Đáp án A