\(\left(2x-3\right)^2+4x^2-9=\left(2x-3\right)\left(3x+5\right)\)
\(\Leftrightarrow\left(2x-3\right)^2+\left(2x-3\right)\left(2x+3\right)-\left(2x-3\right)\left(3x+5\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left[\left(2x-3\right)+\left(2x+3\right)-\left(3x+5\right)\right]=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3+2x+3-3x-5\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=5\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};5\right\}\)
Đúng 3
Bình luận (0)
