Question

Giải phương trình sau

x² ≥ 4

Answer 1

\(x^2\ge4\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\) vậy ......................

Answer 2

\(x^2\ge4\)

\(\Leftrightarrow x^2-4\ge0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\ge0\)

TH 1 : \(\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

TH 2 : \(\left(x-2\right)\left(x+2\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2>0\\x+2>0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x>-2\end{matrix}\right.\)

\(\Leftrightarrow x>2\)

Vậy ...