Question

Giải pt: \(\sqrt{2x^2-9}=-x\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-9=x^2\\x< =0\end{matrix}\right.\Leftrightarrow x=-3\)

Answer 2

`[47]`

`|2x^2 - 9| = x^2`

`->` \(\left[{}\begin{matrix}2x^2-9=x^2\\2x^2-9=-x^2\end{matrix}\right.\)

`->` \(\left[{}\begin{matrix}x^2=9\\3x^2=9\end{matrix}\right.\)

`->` \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\)

Answer 3

\(\sqrt{2x^2-9}=-x\)

\(\Leftrightarrow2x^2-9=x^2\)

\(\Leftrightarrow2x^2-x^2=9\)

\(\Leftrightarrow x^2=9\)

\(\Leftrightarrow x=\pm3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2\left(-3\right)^2}-9=-\left(-3\right)\\\sqrt{2.3^2-9=-3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3=3\\3=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\3=-3\end{matrix}\right.\)

\(\Leftrightarrow x=-3\)