Ta có : \(\left(x-2\right)\left(3x+5\right)=\left(2x-4\right)\left(x+1\right)\)
=> \(\left(x-2\right)\left(3x+5\right)-\left(2x-4\right)\left(x+1\right)=0\)
=> \(\left(x-2\right)\left(3x+5\right)-\left(x-2\right)2\left(x+1\right)=0\)
=> \(\left(x-2\right)\left(3x+5-2\left(x+1\right)\right)=0\)
=> \(\left(x-2\right)\left(3x+5-2x-2\right)=0\)
=> \(\left(x-2\right)\left(x+3\right)=0\)
=> \(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{2,-3\right\}\)
\(\left(x-2\right)\cdot\left(3x+5\right)=\left(2x-4\right)\cdot\left(x+1\right)\\ \Leftrightarrow3x^2+5x-6x-10=2x^2+2x-4x-4\\ \Leftrightarrow3x^2-x-10=2x^2-2x-4\\ \Leftrightarrow3x^2-x-10-2x^2+2x+4=0\\ \Leftrightarrow x^2+x-6=0\\ \Leftrightarrow\left(-3-x\right)\cdot\left(2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3-x=0\\2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Ta có : (x-2)(3x+5)=(2x-4)(x+1)
<=> 3x^2 +5x -6x -10= 2x^2 +2x -4x -4
<=> x^2 +x -6 =0
<=> x^2 -2x +3x -6=0
<=> x(x-2) +3(x-2)=0
<=> (x-2)(x+3)=0
<=> x=2 hoặc x= -3
