giải pt `a, 2x(x-5)-x+5=0` `b, (x+3)^2-(5-x)(x+3)=0` `c, (x+2)(3-4x)=x^2+4x+4`

Đáp án:

`a)`

`2x(x-5)-x+5=0`

`⇔2x(x-5)-(x-5)=0`

`⇔(x-5)(2x-1)=0`

$⇔\left[\begin{matrix}x-5=0\\ 2x-1=0\end{matrix}\right.$

$⇔\left[\begin{matrix}x=5\\ 2x=1\end{matrix}\right.$

$⇔\left[\begin{matrix}x=5\\ x=\dfrac{1}{2}\end{matrix}\right.$

Vậy `S={5;1/2}`

`b)`

`(x+3)^(2)-(5-x)(x+3)=0`

`⇔(x+3)(x+3-5+x)=0`

`⇔(x+3)(2x-2)=0`

`⇔(x+3).2.(x-1)=0`

`⇔(x+3)(x-1)=0`

$⇔\left[\begin{matrix}x+3=0\\ x-1=0\end{matrix}\right.$

$⇔\left[\begin{matrix}x=-3\\ x=1\end{matrix}\right.$

Vậy `S={-3;1}`

`c)`

`(x+2)(3-4x)=x^(2)+4x+4`

`⇔(x+2)(3-4x)=x^(2)+2.x.2+2^2`

`⇔(x+2)(3-4x)-(x+2)^2=0`

`⇔(x+2)(3-4x-x-2)=0`

`⇔(x+2)(-5x+1)=0`

$⇔\left[\begin{matrix}x+2=0\\ -5x+1=0\end{matrix}\right.$

$⇔\left[\begin{matrix}x=-2\\ -5x=-1\end{matrix}\right.$

$⇔\left[\begin{matrix}x=-2\\ x=\dfrac{1}{5}\end{matrix}\right.$

Vậy `S={-2;1/5}`