Toán Lớp 8: Tìm x :
( x – 1 ) ( x + 2 ) + ( x + 2 ) ² = 0
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Toán Lớp 8: Tìm x :
( x – 1 ) ( x + 2 ) + ( x + 2 ) ² = 0
Comments ( 2 )
Giải đáp:
Lời giải và giải thích chi tiết:
$(x-1)(x+2)+(x+2)^2=0$
$\Rightarrow$ $(x+2)(x-1+x+2)=0$
$\Rightarrow$ $(x+2)(2x+1)=0$
$\Rightarrow$ $\left[\begin{matrix} x+2=0\\ 2x+1=0\end{matrix}\right.$
$\Rightarrow$ $\left[\begin{matrix} x=-2\\ 2x=-1\end{matrix}\right.$
$\Rightarrow$ $\left[\begin{matrix} x=-2\\x= \dfrac{-1}{2}\end{matrix}\right.$
Vậy $\left[\begin{matrix} x=-2\\x= \dfrac{-1}{2}\end{matrix}\right.$
Giải đáp:
$(x-1).(x+2)+(x+2)^2=0$
$⇔ (x-1).(x+2)+(x+2).(x+2)=0$
$⇔ (x+2).(x-1+x+2)=0$
$⇔ (x+2).(2x+1)=0$
$⇔$ \(\left[ \begin{array}{l}x+2=0\\2x+1=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=-2\\2x=-1\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=-2\\x=-\dfrac{1}{2}\end{array} \right.\)
$\text{Vậy x ∈ {-2;-1/2}}$