\(\sqrt{2x+15}=32x^2+32x-20\)(1)
ĐK : \(x\ge-\dfrac{15}{2}\)
\(\left(1\right)\Leftrightarrow\sqrt{2x+15}-4=32x^2+32x-24\)
\(\Leftrightarrow\dfrac{2x-1}{\sqrt{2x+15}+4}=\left(2x-1\right)\left(2x+3\right)\)
\(\Leftrightarrow\left(2x-1\right)\left(\dfrac{1}{\sqrt{2x+15}+4}-\left(2x+3\right)\right)=0\)
Làm tiếp nhé!!