\(2f\left(x-2\right)=0\Leftrightarrow f\left(x-2\right)=0\) \(\Rightarrow\left[{}\begin{matrix}x-2=x_1\\x-2=0\\x-2=x_2\end{matrix}\right.\) với \(\left\{{}\begin{matrix}x_1< -2\\x_2>2\end{matrix}\right.\)
\(\Rightarrow x=\left\{x_1+2;2;x_2+2\right\}\) có 3 nghiệm
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