Tìm x biết a,(x-2).(x^2+2x+7) + 2 (x^2-4) – 5(x-2)=0 b, x^3+27+(x+3)(x-9)=0 20/07/2021 Bởi Rylee Tìm x biết a,(x-2).(x^2+2x+7) + 2 (x^2-4) – 5(x-2)=0 b, x^3+27+(x+3)(x-9)=0
`a,(x-2)(x^2+2x+7)+2(x^2-4)-5(x-2)=0` $\Leftrightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0$ $\Leftrightarrow\left(x-2\right)\left(x^2+4x+6\right)=0$ `⇔x-2=0` $\Leftrightarrow x=2$ `b, x^3+27+(x+3)(x-9)=0` $⇔x^3+3^3+\left(x+3\right)\left(x-9\right)=0$ $⇔\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0$ $⇔\left(x+3\right)\left(x^2-3x+9+x-9\right)=0$ $⇔x\cdot\left(x+3\right)\cdot\left(x-2\right)=0$ $\Rightarrow\left\{{}\begin{matrix}x=0\\x+3=0\\x-2=0\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.$ Xin hay nhất ! Bình luận
`a,(x-2)(x^2+2x+7)+2(x^2-4)-5(x-2)=0`
$\Leftrightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0$
$\Leftrightarrow\left(x-2\right)\left(x^2+4x+6\right)=0$
`⇔x-2=0`
$\Leftrightarrow x=2$
`b, x^3+27+(x+3)(x-9)=0`
$⇔x^3+3^3+\left(x+3\right)\left(x-9\right)=0$
$⇔\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0$
$⇔\left(x+3\right)\left(x^2-3x+9+x-9\right)=0$
$⇔x\cdot\left(x+3\right)\cdot\left(x-2\right)=0$
$\Rightarrow\left\{{}\begin{matrix}x=0\\x+3=0\\x-2=0\end{matrix}\right.$
$\Rightarrow\left\{{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.$
Xin hay nhất !
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