Question

Tìm x

x + 1 = \(\left(x+1\right)^2\)

Giúp mk bài này vs mình cám ơn trước ạ

Answer 1

x + 1 =

<=> x + 1 = x\(^2\) + 2x+1

<=> x\(^2\) + 2x+1 - x -1 = 0

<=>x\(^2\) + x = 0

<=>x(x+1)=0

<=> x=0 hoặc x+1 = 0

<=> x=0 hoặc x=-1

Vậy x∈{0;-1}

Answer 2

x + 1 = (x + 1)²

⇔ (x + 1) \(-\) (x + 1)² = 0

⇔ (x + 1)[1 \(-\) (x + 1)] = 0

⇔ (x + 1)(1 \(-\) x + 1) = 0

⇔ (x + 1)(\(- \)x) = 0

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

⇔ \(\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)

Vậy x₁ = \(-\)1; x₂ = 0