Question

Giải phương trình: \(x^2+4\sqrt{x+3}=3x+6\)

Answer 1

ĐKXĐ: \(x\ge-3\)

\(\Leftrightarrow x^2-2x+1-x-3+4\sqrt{x+3}-4=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{x+3}\\\sqrt{x+3}=3-x\left(x\le3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1=x+3\\x+3=9-6x+x^2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2-7x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=6\left(l\right)\end{matrix}\right.\)