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Question

Tìm x:

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

b/$x\left(3x-2\right)-5\left(2-3x\right)=0$

Phong Thần · Accepted Answer

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

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

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

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

b) $x\left(3x-2\right)-5\left(2-3x\right)=0$

$\Rightarrow x\left(3x-2\right)+5\left(3x-2\right)=0$

$\Rightarrow\left(3x-2\right)\left(x+5\right)=0$

$\Rightarrow\left[{}\begin{matrix}3x-2=0\x+5=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}3x=2\x=-5\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\x=-5\end{matrix}\right.$Câu trả lời: a) $x^2\left(x+1\right)+2x\left(x+1\right)=0$