`#3107.101107``x^3 + 1 = x(x + 1)``=> x^3 + 1 - x(x + 1) = 0``=> x^3 + 1 - x^2 - x = 0``=> x^3 - x^2 - x + 1 = 0``=> x^2(x - 1) - (x - 1) = 0``=> (x^2 - 1)(x - 1) = 0``=> (x + 1)(x - 1)(x - 1) = 0``=> (x + 1)(x - 1)^2 = 0``=>`\(\left[{}\begin{matrix}x+1=0\\\left(x-1\right)^2=0\end{matrix}\right.\)`=>`\(\left[{}\begin{matrix}x=-1\\x-1=0\end{matrix}\right.\)`=>`\(\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)Vậy, `x \in {-1; 1}.`Câu trả lời: `#3107.101107``x^3 + 1 = x(x + 1)``=> x^3 + 1 - x(x + 1) = 0``=> x^3 + 1 - x^2 - x = 0``=> x^3 - x^2 - x + 1 = 0``=> x^2(x - 1) - (x - 1) = 0``=> (x^2 - 1)(x - 1) = 0``=> (x + 1)(x - 1)(x - 1) = 0``=> (x + 1)(x - 1)^2 = 0``=>`\(\left[{}\begin{matrix}x+1=0\\\left(x-1\right)^2=0\end{matrix}\right.\)`=>`\(\left[{}\begin{matrix}x=-1\\x-1=0\end{matrix}\right.\)`=>`\(\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)Vậy, `x \in {-1; 1}.`

`x^3 + 1=x(x+1)`` (x+1)(x^2- x+1)-x(x+1)=0`` (x+1)[(x^2-x+1)-x)]=0`` (x+1)(x^2-x+1-x)=0`` (x+1)(x^2-2x+1)=0`` (x+1)(x-1)^2`` ` \(\left[{}\begin{matrix}x+1=0\\\left(x-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\)` ` \(\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\) Vậy `x=-1` hoặc `x=1`Câu trả lời: `x^3 + 1=x(x+1)`` (x+1)(x^2- x+1)-x(x+1)=0`` (x+1)[(x^2-x+1)-x)]=0`` (x+1)(x^2-x+1-x)=0`` (x+1)(x^2-2x+1)=0`` (x+1)(x-1)^2`` ` \(\left[{}\begin{matrix}x+1=0\\\left(x-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\)` ` \(\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\) Vậy `x=-1` hoặc `x=1`