Question

Tìm x

x+1=(x+1)²

Answer 1

Ta có : (x+1)²-(x+1)=0

x^2+2x+1-x-1=0

x^2+x=0

x^2=x

=>x=0,1

Vậy x=1 hoặc x=2

Answer 2

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

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

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

<=> -x² +x - 2x = -1 + 1

<=> -x² - x = 0