Question

Giải phương trình: \(\sqrt{10x+1}+\sqrt{3x-5}=\sqrt{9x+4}+\sqrt{2x-2}\)

Answer 1

ĐKXĐ: \(x\ge\dfrac{5}{3}\)

\(\sqrt{10x+1}-\sqrt{9x+4}+\sqrt{3x-5}-\sqrt{2x-2}=0\)

\(\Leftrightarrow\dfrac{x-3}{\sqrt{10x+1}+\sqrt{9x+4}}+\dfrac{x-3}{\sqrt{3x-5}+\sqrt{2x-2}}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{\sqrt{10x+1}+\sqrt{9x+4}}+\dfrac{1}{\sqrt{3x-5}+\sqrt{2x-2}}\right)=0\)

\(\Leftrightarrow x-3=0\Rightarrow x=3\)

Do \(\dfrac{1}{\sqrt{10x+1}+\sqrt{9x+4}}+\dfrac{1}{\sqrt{3x-5}+\sqrt{2x-2}}>0\) \(\forall x\ge\dfrac{5}{3}\)

Vậy pt có nghiệm duy nhất \(x=3\)

Answer 2

@Nguyễn Việt Lâm @TRẦN MINH HOÀNG @ Mashiro Shiina