Question

\(x^2+3x-\dfrac{7}{4}=2\sqrt{2x-3}\)

Answer 1

Ta có: \(x^2+3x-\dfrac{7}{4}=2\sqrt{2x-3}\) ( ĐK: \(x\ge\dfrac{3}{2}\) )

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

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

<=> \(\left\{{}\begin{matrix}x+\dfrac{1}{2}=0\\\sqrt{2x-3}=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\2x-3=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\x=2\end{matrix}\right.\)

=> PT vô nghiệm

~~~~~ Là lá la , vô nghiệm kìa ~~~~~~~