Question

Giải pt: \(x^2+2\sqrt{2x+7}=2\sqrt{-2x+3}+5\)

Answer 1

Ta có \(x^2+2\sqrt{2x+7}=2\sqrt{-2x+3}+5\)ĐKXĐ \(-\frac{7}{2}\le x\le\frac{3}{2}\)

<=> \(\left(x^2+2x-3\right)+\left(-x-5+2\sqrt{2x+7}\right)+\left(3-x-2\sqrt{-2x+3}\right)=0\)

<=>\(x^2+2x-3+\frac{-x^2-2x+3}{x+5+2\sqrt{2x+7}}+\frac{x^2+2x-3}{3-x+2\sqrt{-2x+3}}=0\)

<=> \(\orbr{\begin{cases}x^2+2x-3=0\\1-\frac{1}{x+5+2\sqrt{2x+7}}+\frac{1}{3-x+2\sqrt{-2x+3}}=0\left(2\right)\end{cases}}\)

Với ĐK \(x\ge-\frac{7}{2}\)

=> \(\frac{1}{x+5+2\sqrt{2x+7}}< 1\)=> phương trinh (2) vô nghiệm

Vậy \(S=\left\{-3;1\right\}\)