Lời giải:

$G=\frac{x^2+x+2}{2x^2-2x+3}$

$\Rightarrow G(2x^2-2x+3)=x^2+x+2$
$\Leftrightarrow x^2(2G-1)-x(2G+1)+(3G-2)=0(*)$

Vì $G$ tồn tại nên dấu "=" tồn tại, điều này có nghĩa là $(*)$ luôn có nghiệm.

$\Rightarrow \Delta=(2G+1)^2-4(2G-1)(3G-2)\geq 0$

$\Leftrightarrow -20G^2+32G-7\geq 0$

$\Leftrightarrow 20G^2-32G+7\leq 0$

$\Leftrightarrow \frac{16+\sqrt{116}}{20}\geq G\geq \frac{16-\sqrt{116}}{20}$

Vậy....

\(C=\frac{4x^2+2x-2}{2\left(x^2-2x+2\right)}=\frac{9\left(x^2-2x+2\right)-5x^2+20x-20}{2\left(x^2-2x+2\right)}=\frac{9}{2}-\frac{5\left(x-2\right)^2}{2\left(x-1\right)^2+2}\le\frac{9}{2}\)

\(C_{max}=\frac{9}{2}\) khi \(x=2\)

\(C=\frac{4x^2+2x-2}{2\left(x^2-2x+2\right)}=\frac{-\left(x^2-2x+2\right)+5x^2}{2\left(x^2-2x+2\right)}=-\frac{1}{2}+\frac{5x^2}{2\left(x-1\right)^2+2}\ge-\frac{1}{2}\)

\(C_{min}=-\frac{1}{2}\) khi \(x=0\)

Câu D bạn coi lại đềm kết quả rất xấu: \(\frac{3-\sqrt{17}}{12}\le D\le\frac{3+\sqrt{17}}{12}\)

a, \(3x^3-5x^2-x-2>0\)

\(< =>3x^3+x^2+x-6x^2-2x-2>0\)

\(< =>x\left(3x^2+x+1\right)-2\left(3x^2+x+1\right)>0\)

\(< =>\left(x-2\right)\left(3x^2+x+1\right)>0\)

có \(3x^2+x+1=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{3}\right)=3\left[x^2+2.\dfrac{1}{6}x+\dfrac{1}{36}+\dfrac{35}{36}\right]\)

\(=3\left[\left(x+\dfrac{1}{6}\right)^2+\dfrac{35}{36}\right]>0=>x-2>0< =>x>2\)

b, \(A=2x^2+y^2-2xy-2x+3\)

\(A=x^2-2xy+y^2+x^2-2x+1+2\)

\(A=\left(x-y\right)^2+\left(x-1\right)^2+2\ge2\)

dấu"=" xảy ra<=>\(x=y=1\)

1/ \(x^2-2x+7\)

\(=x^2-\frac{1}{2}\cdot2x+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+7\)

\(=x^2-\frac{1}{2}\cdot2x+\left(\frac{1}{2}\right)^2-\frac{1}{4}+7\)

\(=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}+7\)

\(=\left(x-\frac{1}{2}\right)^2+\frac{27}{4}\)

Có  \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{27}{4}\ge\frac{27}{4}\)

\(\Rightarrow GTNNx^2-2x+7=\frac{27}{4}\)

               với  \(\left(x-\frac{1}{2}\right)^2=0;x=\frac{1}{2}\)

2/ \(4x^2+2x+9\)

\(=\left(2x\right)^2+2\cdot2\cdot\frac{1}{2}x+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+9\)

\(=\left(2x+\frac{1}{2}\right)^2-\frac{1}{4}+9\)

\(=\left(2x+\frac{1}{2}\right)^2+\frac{35}{4}\)

có \(\left(2x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(2x+\frac{1}{2}\right)^2+\frac{35}{4}\ge\frac{35}{4}\)

\(\Rightarrow GTNN4x^2+2x+9=\frac{35}{4}\)

                với  \(\left(2x+\frac{1}{2}\right)^2=0;x=-\frac{1}{4}\)