Question

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

Answer 1

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

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

\(\Leftrightarrow x^2+3x+3\sqrt{x^2+3x}-10=0\) (1)

đặc \(\sqrt{x^2+3x}=t\left(t\ge0\right)\)

\(\Rightarrow\left(1\right)\Leftrightarrow t^2+3t-10=0\Leftrightarrow\left[{}\begin{matrix}t=2\left(N\right)\\t=-5\left(L\right)\end{matrix}\right.\)

\(t=2\Leftrightarrow\sqrt{x^2+3x}=2\Leftrightarrow x^2+3x-4=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

vậy \(x=1;x=-4\)