Question

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

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}2x+7>=0\\-2x+3>=0\end{matrix}\right.\Leftrightarrow-\dfrac{7}{2}< =x< =\dfrac{3}{2}\)

PT\(\Leftrightarrow x^2-1+2\sqrt{2x+7}=2\sqrt{-2x+3}+4\)

=>\(\left(x-1\right)\left(x+1\right)+2\sqrt{2x+7}-6=2\sqrt{-2x+3}-2\)

=>\(\left(x-1\right)\left(x+1\right)+2\cdot\dfrac{2x+7-9}{\sqrt{2x+7}+3}=2\cdot\dfrac{-2x+3-1}{\sqrt{-2x+3}+1}\)

=>\(\left(x-1\right)\left(x+1\right)+\dfrac{4\left(x-1\right)}{\sqrt{2x+7}+3}-2\cdot\dfrac{-2\left(x-1\right)}{\sqrt{-2x+3}+1}=0\)

=>\(\left(x-1\right)\left(x+1+\dfrac{4}{\sqrt{2x+7}+3}+\dfrac{4}{\sqrt{-2x+3}+1}\right)=0\)

=>x-1=0

=>x=1(nhận)