\(\left|x-\dfrac{3}{4}\right|+2x^2+\left|2x^2+2,5\right|\)
\(\Rightarrow\left|x-\dfrac{3}{4}\right|+2x^2=2x^2+2,5\) (vì \(2x^2+2,5>0\forall x\))
\(\Rightarrow\left|x-\dfrac{3}{4}\right|=2,5\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=2,5\\x-\dfrac{3}{4}=-2,5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{4}\\x=-\dfrac{7}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{13}{4};-\dfrac{7}{4}\right\}\).
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