Question

Giải phương trình:

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

Answer 1

\(x\ge\dfrac{2}{3}\)

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

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

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

Xét (1): \(VT=x+1>1\) \(\forall x\ge\dfrac{2}{3}\)

\(\sqrt{3x-2}+\sqrt{x+1}>1\) \(\forall x\ge\dfrac{2}{3}\)

\(\Rightarrow VP=\dfrac{1}{\sqrt{3x-2}+\sqrt{x+1}}< 1\) \(\forall x\ge\dfrac{2}{3}\)

\(\Rightarrow VT>VP\Rightarrow\) \(\) pt (1) vô nghiệm

Vậy pt có nghiệm duy nhất \(x=\dfrac{3}{2}\)