Vì \(x^2+1\ne0\) nên ta có thể viết lại:

\(\left(x^2+1\right)Q=2x^2+2x+2\Leftrightarrow Qx^2+Q=2x^2+2x+2\)\(\Leftrightarrow Qx^2-2x^2-2x+Q-2=0\Leftrightarrow\left(Q-2\right)x^2-2x+Q-2=0\) (*)

pt (*) có nghiệm khi \(\Delta'=\left(-1\right)^2-\left(Q-2\right)\left(Q-2\right)=1-\left(Q-2\right)^2\ge0\)\(\Leftrightarrow\left(Q-2\right)^2\le1\)\(\Leftrightarrow-1\le Q-2\le1\)\(\Leftrightarrow1\le Q\le3\) (đpcm)

khó vl

\(y=\left(x+2\right)\left(3-x\right)\)

\(=3x-x^2+6-2x\)

\(=-x^2+x+6\)

=>y'=-2x+1

Đặt y'=0

=>-2x+1=0

=>-2x=-1

=>\(x=\dfrac{1}{2}\)

\(f\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}+2\right)\left(3-\dfrac{1}{2}\right)=\dfrac{5}{2}\cdot\dfrac{5}{2}=\dfrac{25}{4}\)

\(f\left(-2\right)=\left(-2+2\right)\left(3+2\right)=0\)

\(f\left(3\right)=\left(3+2\right)\left(3-3\right)=0\)

=>\(y_{max\left[-2;3\right]}=\dfrac{25}{4}\)

Chọn C

\(\frac{1}{3}< =\frac{x^2+x+1}{x^2-x+1}\Rightarrow x^2-x+1< =3x^2+3x+3\Rightarrow x^2-x+1-3x^2-3x-3< =0\)

\(\Rightarrow-2x^2-4x-2< =0\Rightarrow-2\left(x^2+2x+1\right)< =0\Rightarrow-2\left(x+1\right)^2< =0\)

vì \(\left(x+1\right)^2>=0;-2< 0\Rightarrow-2\left(x+1\right)^2< =0\)luôn đúng \(\Rightarrow\frac{1}{3}< =\frac{x^2+x+1}{x^2-x+1}\)luôn dúng (1)

cái kia cx tương tự như vậy nhé

Ta có:

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

Ta có:\(2+\frac{2x}{x^2+1}-1=1+\frac{2x}{x^2+1}\)

\(=\frac{x^2+2x+1}{x^2+1}=\frac{\left(x+1\right)^2}{x^2+1}\ge0\)  \(\Rightarrow\frac{2.\left(x^2+x+1\right)}{x^2+1}\ge1\)

\(2+\frac{2x}{x^2+1}-3=\frac{2x}{x^2+1}-1=\frac{-x^2+2x-1}{x^2+1}\)

\(=\frac{-\left(x-1\right)^2}{x^2+1}\le0\) \(\Rightarrow\frac{2.\left(x^2+x+1\right)}{x^2+1}\le3\)

Vậy \(1\le\frac{2.\left(x^2+x+1\right)}{x^2+1}\le3\)

\(\cdot\left(x+1\right)^2\ge0\)

\(\Rightarrow x^2+2x+1>0\)

\(\Rightarrow2x^2+4x+2\ge0\)

 \(\Rightarrow\left(3x^2+3x+3\right)-\left(x^2-x+1\right)\ge0\)

\(\Rightarrow3\left(x^2+x+1\right)\ge x^2-x+1\)

\(\Rightarrow\)\(\frac{x^2+x+1}{x^2-x+1}\ge\frac{1}{3}\) (1)

\(\cdot\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow2x^2-4x+2\ge0\)

\(\Rightarrow\left(3x^2-3x+3\right)-\left(x^2+x+1\right)\ge0\)

\(\Rightarrow3\left(x^2-x+1\right)\ge x^2+x+1\)

\(\Rightarrow\frac{x^2+x+1}{x^2-x+1}\le3\)(2)

Từ(1),(2) => đpcm

\(\dfrac{x^2+x+1}{x^2-x+1}-\dfrac{1}{3}=\dfrac{3x^2+3x+3-x^2+x-1}{3\left(x^2-x+1\right)}\)

\(=\dfrac{2x^2+4x+2}{3\left(x^2-x+1\right)}=\dfrac{2\left(x+1\right)^2}{3\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}}\ge0\)

Do đó: \(\dfrac{1}{3}\le\dfrac{x^2+x+1}{x^2-x+1}\)(1)

\(\dfrac{x^2+x+1}{x^2-x+1}-3=\dfrac{x^2+x+1-3x^2+3x-3}{x^2-x+1}\)

\(=\dfrac{-2x^2+4x-2}{x^2-x+1}=\dfrac{-2\left(x-1\right)^2}{x^2-x+1}\le0\)

Do đó: \(\dfrac{x^2+x+1}{x^2-x+1}\le3\)(2)

Từ (1)và (2) suy ra ĐPCM