Question

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

Answer 1

ĐKXĐ: \(2\le x\le4\)

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

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

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

Ta thấy: (1) \(\Leftrightarrow\frac{1}{\sqrt{x+2}+1}=\frac{1}{\sqrt{4-x}+1}+2x+1\)

Với \(2\le x\le4\) ta có: \(\frac{1}{\sqrt{x+2}+1}< \frac{1}{3}\); \(\frac{1}{\sqrt{4-x}+1}+2x+1>2x+1\ge5\).

Do đó (1) vô nghiệm.

Vậy nghiệm của pt là x = 3.