Question

(x + 1).(x + 2) - (x - 3)-(x + 4) = 6

Giúp vs ạ đg cần gấp

 

Answer 1

\((x+1)(x+2)-(x-3)-(x+4)=6\)

\(\Leftrightarrow x^2 +3x+2-x^2-x+12=6\)

\(\Leftrightarrow 2x=-8\)

\(\Leftrightarrow x=-4\)

Answer 2

`(x + 1).(x + 2) - (x - 3)-(x + 4) = 6`

`<=> x^2 + 3x +2 -x +3 -x-4 =6`

`<=> x^2 - x - 1  =6`

`<=>x^2 -x -7=0`

Pt có : \(\Delta=b^2-4ac=\left(-1\right)^2-4.\left(-7\right)=29>0\)

=> Pt có 2 nghiệm phân biệt

\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{1+\sqrt{29}}{2}\)

\(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{1-\sqrt{29}}{2}\)

Vậy \(S=\left\{\dfrac{1\pm\sqrt{29}}{2}\right\}\)