\(\left(x+1\right)^2=\left(x+1\right)^3\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)^3=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(1-x-1\right)=0\)
\(\Leftrightarrow-x\left(x+1\right)^2=0\)
\(\Leftrightarrow-x=0\)
\(\Leftrightarrow x=0\)
Vậy \(S=\left\{0\right\}\)
<=> \(\left(x+1\right)\left(x+1\right)=\left(x+1\right)\left(x+1\right)\left(x+1\right)\)
mà : \(\left(x+1\right)\left(x+1\right)=\left(x+1\right)\left(x+1\right)\)
=> \(\left(x+1\right)\left(x+1\right)< \left(x+1\right)\left(x+1\right)\left(x+1\right)\) ( luôn đúng)
=> pt vô nghiệm
= (x + 1)2 - (x + 1)3 => 0
= (x + 1)2 (1 - x - 1) => 0
= - x(x + 1)2 => 0
= - x => 0
= x = 0
vậy S = [ 0 ]
