Question

giải phương trình:

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

Answer 1

ĐKXĐ: ...

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

Đặt \(\sqrt{2x+3}=t\ge0\)

\(\Rightarrow x^2+4x-t^2-4t=0\)

\(\Leftrightarrow\left(x+t\right)\left(x-t\right)+4\left(x-t\right)=0\)

\(\Leftrightarrow\left(x-t\right)\left(x+t+4\right)=0\)

\(\Leftrightarrow x=t\) (do \(x\ge-\frac{3}{2}\Rightarrow x+4+t>0\))

\(\Leftrightarrow\sqrt{2x+3}=x\left(x\ge0\right)\)

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