Question

Giải pt:  \(\sqrt{4x^2-9}=2\sqrt{2x+3}\)

Answer 1

\(\sqrt{4x^2-9}=2\sqrt{2x+3}\left(ĐK:4x^2-9>=0;2x+3>=0\right)\\ < =>4x^2-9=4\left(2x+3\right)\\ < =>\left(2x+3\right)\left(2x-3\right)-4\left(2x+3\right)=0\\ < =>\left(2x+3\right)\left(2x-3-4\right)=0\\ < =>\left(2x+3\right)\left(2x-7\right)=0\\ =>\left[{}\begin{matrix}2x+3=0\\2x-7=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(TMDK\right)\\x=\dfrac{7}{2}\left(TMDK\right)\end{matrix}\right.=>S=\left\{-\dfrac{3}{2};\dfrac{7}{2}\right\}\)