a) \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)
b) \(x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)
c) \(x^2+y^2-x-6y+10=\left(x-\dfrac{1}{2}\right)^2+\left(y-3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(ĐTXR\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=3\end{matrix}\right.\)
d) \(\dfrac{2}{6x-5-9x^2}=\dfrac{2}{-\left(3x-1\right)^2-4}\)
Do \(-\left(3x-1\right)^2-4\le-4\Rightarrow\dfrac{2}{-\left(3x-1\right)^2-4}\ge\dfrac{2}{-4}=-\dfrac{1}{2}\)
\(ĐTXR\Leftrightarrow x=\dfrac{1}{3}\)
e) \(\dfrac{3}{2x^2+2x+3}=\dfrac{3}{2\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{2}}\)
Do \(2\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{2}\ge\dfrac{5}{2}\Rightarrow\dfrac{3}{2\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{2}}\le\dfrac{6}{5}\)
\(ĐTXR\Leftrightarrow x=-\dfrac{1}{2}\)
