Question

giải pt : \(\sqrt{x-2}+\sqrt{4-x}+\sqrt{2x-5}=2x^2-5x\)

Answer 1

ĐKXĐ: \(\frac{5}{2}\le x\le4\)

\(\Leftrightarrow\sqrt{x-2}-1+\sqrt{2x-5}-1=2x^2-5x-3+1-\sqrt{4-x}\)

\(\Leftrightarrow\frac{x-3}{\sqrt{x-2}+1}+\frac{2\left(x-3\right)}{\sqrt{2x-5}+1}=\left(2x+1\right)\left(x-3\right)+\frac{x-3}{1+\sqrt{4-x}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\frac{1}{\sqrt{x-2}+1}+\frac{2}{\sqrt{2x-5}+1}=2x+1+\frac{1}{1+\sqrt{4-x}}\left(1\right)\end{matrix}\right.\)

Xét (1), ta có \(VT< 3\) , mà \(x\ge\frac{5}{2}\Rightarrow2x+1>6>3\Rightarrow VP>3\)

Vậy (1) vô nghiệm hay pt có nghiệm duy nhất \(x=3\)