1) \(A=2x^2+9x-13=2\left(x+\dfrac{9}{4}\right)^2-\dfrac{185}{8}\ge-\dfrac{185}{8}\forall x\)
=> \(A_{min}=-\dfrac{185}{5}\Leftrightarrow x=-\dfrac{9}{4}\)
2) \(B=\dfrac{3x^2-1}{x^2+2}\)\(=\dfrac{3\left(x^2+2\right)-7}{x^2+2}=3-\dfrac{7}{x^2+2}\)
Có \(x^2+2\ge2\) \(\Leftrightarrow\dfrac{7}{x^2+2}\le\dfrac{7}{2}\)\(\Leftrightarrow-\dfrac{7}{x^2+2}\ge-\dfrac{7}{2}\) \(\Leftrightarrow3-\dfrac{7}{x^2+2}\ge-\dfrac{1}{2}\)
\(\Rightarrow B_{min}=-\dfrac{1}{2}\Leftrightarrow x=0\)
3) \(C=\dfrac{x^2-4x-4}{x^2-4x+5}\)\(=\dfrac{\left(x^2-4x+5\right)-9}{x^2-4x+5}=1-\dfrac{9}{x^2-4x+5}\)
Có \(x^2-4x+5=\left(x-2\right)^2+1\ge1\) \(\Rightarrow\dfrac{9}{x^2-4x+5}\le9\) \(\Leftrightarrow-\dfrac{9}{x^2-4x+5}\ge-9\)
\(\Leftrightarrow1-\dfrac{9}{x^2-4x+5}\ge-8\)
\(\Rightarrow C_{min}=-8\Leftrightarrow x=2\)
4) \(D=\dfrac{-x^2+x-10}{x^2-2x+1}\) (đk:\(x\ne1\))
Xét \(40\left(-x^2+x-10\right)+39\left(x^2-2x+1\right)=-x^2-38x-361\)
\(=-\left(x+19\right)^2\le0\)
\(\Rightarrow40\left(-x^2+x-10\right)\le-39\left(x^2-2x+1\right)\)
\(\Leftrightarrow\dfrac{-x^2+x-10}{x^2-2x+1}\le-\dfrac{39}{40}\)
\(\Leftrightarrow D\le-\dfrac{39}{40}\) => \(D_{max}=-\dfrac{39}{40}\Leftrightarrow x=-19\)
5)\(E=\dfrac{8x+12}{x^2+4}=\dfrac{x^2+8x+16-\left(x^2+4\right)}{x^2+4}\)\(=\dfrac{\left(x+4\right)^2}{x^2+4}-1\ge-1\)
\(\Rightarrow E_{min}=-1\Leftrightarrow x=-4\)
\(E=\dfrac{8x+12}{x^2+4}=\dfrac{4\left(x^2+4\right)-\left(4x^2-8x+4\right)}{x^2+4}\)\(=4-\dfrac{4\left(x-1\right)^2}{x^2+4}\le4\)
\(\Rightarrow E_{max}=4\Leftrightarrow x=1\)
