Câu hỏi của Võ Thị Bi

Question

tìm x biết

a) x2(x+1)+2x(x+1)=0

Bùi Mạnh Khôi · Accepted Answer

$x^2\left(x+1\right)+2x\left(x+1\right)=0$

$\Leftrightarrow\left(x^2+2x\right)\left(x+1\right)=0$

$\Leftrightarrow x\left(x+2\right)\left(x+1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x=0\x+2=0\x+1=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=0\x=-2\x=-1\end{matrix}\right.$

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

$\Leftrightarrow\left(x^2+2x\right)\left(x+1\right)=0$

$\Leftrightarrow x\left(x+2\right)\left(x+1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x=0\x+2=0\x+1=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=0\x=-2\x=-1\end{matrix}\right.$

Vậy $\left[{}\begin{matrix}x=0\x=-2\x=-1\end{matrix}\right.$

Dũng Nguyễn · Answer

Ta có:$x^2\left(x+1\right)+2x\left(x+1\right)=0$

$\Rightarrow\left(x^2+2x\right)\left(x+1\right)=0$

$\Rightarrow\left[{}\begin{matrix}x^2+2x=0\x+1=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x\left(x+2\right)=0\x=-1\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\x+2=0\end{matrix}\right.\x=-1\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\x=-2\end{matrix}\right.\x=-1\end{matrix}\right.$

Vậy $x\in\left\{0;-1;-2\right\}$Câu trả lời: Ta có:$x^2\left(x+1\right)+2x\left(x+1\right)=0$

$\Rightarrow\left(x^2+2x\right)\left(x+1\right)=0$

$\Rightarrow\left[{}\begin{matrix}x^2+2x=0\x+1=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x\left(x+2\right)=0\x=-1\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\x+2=0\end{matrix}\right.\x=-1\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\x=-2\end{matrix}\right.\x=-1\end{matrix}\right.$

Vậy $x\in\left\{0;-1;-2\right\}$

Bài 5: Những hằng đẳng thức đáng nhớ (Tiếp)