g p t `x ( x + 1 ) + x( x - 5 ) = 2 ( 2 - x ) `

Answer

`x . (x + 1) + x . (x - 5) = 2 . (2 - x)`

`<=> x . x + x . 1 + x . x - x . 5 = 2 . 2 - 2 . x`

`<=> x^2 + x + x^2 - 5x = 4 - 2x`

`<=> x^2 + x + x^2 - 5x - 4 + 2x = 0`

`<=> (x^2 + x^2) + (x - 5x + 2x) - 4 = 0`

`<=> 2x^2 - 2x - 4 = 0`

`<=> 2 . (x^2 - x - 2) = 0`

`<=> 2 . (x^2 - 2x + x - 2) = 0`

`<=> 2 . [(x^2 - 2x) + (x - 2)] = 0`

`<=> 2 . [x . (x - 2) + (x - 2) . 1] = 0`

`<=> 2 . (x + 1) . (x - 2) = 0`

`<=>` $\left[\begin{matrix} x + 1 = 0\\ x - 2 = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = 0 - 1\\ x = 0 + 2\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = -1\\ x = 2\end{matrix}\right.$

Vậy `S = {-1 ; 2}`

`x(x + 1) + x(x - 5) = 2(2 - x)`

`⇔ x^2 + x + x^2 - 5x  = 4 - 2x`

`⇔ x^2 + x + x^2 - 5x - 4 + 2x = 0`

`⇔ (x^2 + x^2) + (x - 5x + 2x) - 4 = 0`

`⇔ 2x^2 - 2x - 4 = 0`

`⇔ 2(x^2 - x - 2) = 0`

`⇔ 2(x^2 + x - 2x - 2) = 0`

`⇔ 2[(x^2 + x) - (2x + 2)] = 0`

`⇔ 2[x(x + 1) - 2(x + 1)] = 0`

`⇔ 2.(x + 1).(x - 2) = 0`

`⇔` \(\left[ \begin{array}{l}x+1=0\\x-2=0\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=-1\\x=2\end{array} \right.\) 

Vậy pt có tập nghiệm `S = {-1; 2}`