Câu hỏi của Tiểu Bảo Bảo

Question

Giải phương trình $x^2+4\sqrt{x+3}=3x+6$

Phương Anh · Accepted Answer

$x^2+4\sqrt{x+3}=3x+6$

$\Leftrightarrow\left(x^2-2x+1\right)-\left[\left(x+3\right)-4\sqrt{x+3}+4\right]=0$

$\Leftrightarrow\left(x-1\right)^2-\left(\sqrt{x+3}-2\right)^2=0$

$\Leftrightarrow\left(x+\sqrt{x+3}-3\right)\left(x-\sqrt{x+3}+1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=3-x\\sqrt{x+3}=x+1\end{matrix}\right.$

To be continued. . .

$S=\left\{1\right\}$Câu trả lời: $x^2+4\sqrt{x+3}=3x+6$

$\Leftrightarrow\left(x^2-2x+1\right)-\left[\left(x+3\right)-4\sqrt{x+3}+4\right]=0$

$\Leftrightarrow\left(x-1\right)^2-\left(\sqrt{x+3}-2\right)^2=0$

$\Leftrightarrow\left(x+\sqrt{x+3}-3\right)\left(x-\sqrt{x+3}+1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=3-x\\sqrt{x+3}=x+1\end{matrix}\right.$

To be continued. . .

$S=\left\{1\right\}$

Bài 1: Căn bậc hai