Question

Giải Phương Trình: (x+1)\(\sqrt{2x^2-2x}\) = 2x²-3x-2

Answer 1

ĐKXĐ: ...

\(\Leftrightarrow2x^2-2x-\left(x+1\right)\sqrt{2x^2-2x}-x-2=0\)

Đặt \(\sqrt{2x^2-2x}=t\ge0\)

\(\Rightarrow t^2-\left(x+1\right)t-x-2=0\)

\(\Delta=\left(x+1\right)^2+4\left(x+2\right)=\left(x+3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\frac{x+1+x+3}{2}=x+2\\t=\frac{x+1-x-3}{2}=-1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x^2-2x}=x+2\left(x\ge-2\right)\)

\(\Leftrightarrow2x^2-2x=x^2+4x+4\)

\(\Leftrightarrow x^2-6x-4=0\) (casio)