Question

Giải phương trình 1).(x+5)(2-x)=3 căn x^2+3x

 

Answer 1

ĐKXĐ: \(\left[{}\begin{matrix}x\ge0\\x\le-3\end{matrix}\right.\)

\(\left(x+5\right)\left(2-x\right)=3\sqrt{x^2+3x}\)

\(\Leftrightarrow-x^2-3x+10=3\sqrt{x^2+3x}\)

\(\Leftrightarrow x^2+3x+3\sqrt{x^2+3x}-10=0\)

Đặt \(\sqrt{x^2+3x}=t\ge0\)

\(\Rightarrow t^2+3t-10=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-5\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2+3x}=2\)

\(\Leftrightarrow x^2+3x=4\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)