\(\left(5-x\right)\left(2+3x\right)=4-9x^2\)
\(\Leftrightarrow\left(5-x\right)\left(2+3x\right)=\left(2-3x\right)\left(2+3x\right)\)
\(\Leftrightarrow\left(5-x\right)\left(2+3x\right)-\left(2-3x\right)\left(2+3x\right)=0\)
\(\Leftrightarrow\left(2+3x\right)\left(5-x-2+3x\right)=0\)
\(\Leftrightarrow\left(2+3x\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2+3x=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{-\dfrac{2}{3};-\dfrac{3}{2}\right\}\)
Phương trình trên tương đương:
(5-x)(2+3x)=(2-3x)(2+3x)
(5-x)(2+3x)-(2-3x)(2+3x)=0
Đặt 2+3x làm nhân tử chung rồi giải pt tích rồi kết luận
\(\left(5-x\right)\left(2+3x\right)=4-9x^2\)
\(< =>\left(5-x\right)\left(2+3x\right)=2^2-\left(3x\right)^2\)
\(< =>\left(5-x\right)\left(2+3x\right)-\left(2-3x\right)\left(2+3x\right)=0\)
\(< =>\left(2+3x\right)\left(5-x-2+3x\right)=0\)
\(< =>\left(2+3x\right)\left(2x+3\right)=0\)
\(< =>\left[{}\begin{matrix}2+3x=0\\2x+3=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
S=\(\left\{\dfrac{-2}{3},\dfrac{-3}{2}\right\}\)
