\(\sqrt{2x^2-4x+9}=3\)
Ta có: \(2x^2-4x+9=2x^2-4x+2+7=2\left(x^2-2x+1\right)+7=2\left(x+1\right)^2+7>0\forall x\)
Vậy \(\sqrt{2x^2-4x+9}=3\) luôn có nghĩ với mọi x
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\(\sqrt{2x^2-4x+9}=3\)
\(\Leftrightarrow2x^2-4x+9=3^2\)
\(\Leftrightarrow2x^2-4x+9=9\)
\(\Leftrightarrow2x^2-4x=0\)
\(\Leftrightarrow2x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy: ...
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