Question

Gải phương trình: \(\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-1\)

Answer 1

Lời giải:

ĐK: \(2\leq x\le 4\)

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

\(\Leftrightarrow 2x(x-3)+(x-3)-(\sqrt{x-2}-1)-(\sqrt{4-x}-1)=0\)

\(\Leftrightarrow (2x+1)(x-3)-\frac{x-3}{\sqrt{x-2}+1}-\frac{3-x}{\sqrt{4-x}+1}=0\)

\(\Leftrightarrow (x-3)\left(2x+1-\frac{1}{\sqrt{x-2}+1}+\frac{1}{\sqrt{4-x}+1}\right)=0(*)\)

Ta thấy \(\forall 2\leq x\leq 4: 2x+1\geq 5\) và \(\frac{1}{\sqrt{x-2}+1}\leq 1; \frac{1}{\sqrt{4-x}+1}>0\)

\(\Rightarrow 2x+1-\frac{1}{\sqrt{2x-1}+1}+\frac{1}{\sqrt{4-x}+1}>0\)

Do đó từ \((*)\Rightarrow x-3=0\Rightarrow x=3\)