`2(x-3)^2=(x-3)(x+5)^2`
`<=> 2(x-3)^2-(x-3)(x+5)^2=0`
`<=> (x-3)[2(x-3)-(x+5)^2]=0`
`<=> (x-3)(2x-6-x^2-10x-25)=0`
`<=>(x-3)(-x^2-8x-31)=0`
`<=> x=3`
(Vì `-x^2-8x-31=-(x^2+8x+31)=-(x^2+2.x.4+4^2+15)=-(x+4)^2-15 \ne 0 \forall x)`
Vậy `x=3`.
.
`4(x-3)=2x-5(2x+3)`
`<=>4x-12=2x-10x-15`
`<=>4x-12=-8x-15`
`<=>12x=-3`
`<=>x=-1/4`
Vậy `x=-1/4`.
1) \(2.\left(x-3\right).2=\left(x-3\right)\left(x+5\right)\)
\(\Leftrightarrow4x-12=x^2+5x-3x-15\)
\(\Leftrightarrow4x-12-x^2-5x+3x+15=0\)
\(\Leftrightarrow-x^2+2x+3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy: S = {3; -1}
2) \(4\left(x-3\right)=2x-5.\left(2x+3\right)\)
\(\Leftrightarrow4x-12=2x-10x-15\)
\(\Leftrightarrow4x-12-2x+10x+15=0\)
\(\Leftrightarrow12x+3=0\)
<=> 12x = -3
<=> x = -4
Vậy S = {-4}