Question

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

Answer 1

Lời giải:
ĐK: \(x\geq 2\)

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

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

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

Ta thấy vì $x\geq 2$ nên \(\frac{x-1}{\sqrt{2x-1}+\sqrt{x+3}}>0(1)\)

\(\sqrt{x+1}+\sqrt{x-2}< \sqrt{2x-1}+\sqrt{x+3}, \forall x\geq 2\)

\(\Rightarrow \frac{3}{\sqrt{x+1}+\sqrt{x-2}}> \frac{3}{\sqrt{2x-1}+\sqrt{x+3}}\Rightarrow \frac{3}{\sqrt{x+1}+\sqrt{x-2}}-\frac{3}{\sqrt{2x-1}+\sqrt{x+3}}>0(2)\)

Từ $(1),(2)$ suy ra pt $(*)$ vô nghiệm.

Vậy pt vô nghiệm.