\(x+x^2-2=0\)
\(\Leftrightarrow x+x^2-1-1=0\)
\(\Leftrightarrow x-1+x^2-1=0\)
\(\Leftrightarrow\left(x-1\right)+\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(1+x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2+x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\2+x=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-2\end{array}\right.\)
Vậy \(x\in\left\{1;-2\right\}\)
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