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

a)x²+2x+4+1=(x+1)²+1

ma (x+1)² >0

nen (x+1)²+1>1

vay x²+2x+5 min la 1 khi x=-1

\(A=\left(2x^2+3\right)-7\)

\(A=2x^2+3-7\)

\(A=2x^2-4\ge-4\)

vậy Min A=-4 khi và chỉ khi x=0

\(y=2sin^2x+3sinx.cosx+cos^2x\)

\(=-\left(1-2sin^2x\right)+\dfrac{3}{2}sin2x+\dfrac{1}{2}\left(2cos^2x-1\right)+\dfrac{1}{2}\)

\(=-cos2x+\dfrac{3}{2}sin2x+\dfrac{1}{2}cos2x+\dfrac{1}{2}\)

\(=\dfrac{3}{2}sin2x-\dfrac{1}{2}cos2x+\dfrac{1}{2}\)

\(=\dfrac{\sqrt{10}}{2}\left(\dfrac{3}{\sqrt{10}}sin2x-\dfrac{1}{\sqrt{10}}cos2x\right)+\dfrac{1}{2}\)

\(=\dfrac{\sqrt{10}}{2}sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)+\dfrac{1}{2}\)

Vì \(sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)\in\left[-1;1\right]\)

\(\Rightarrow y=\dfrac{\sqrt{10}}{2}sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)+\dfrac{1}{2}\in\left[-\dfrac{\sqrt{10}}{2}+\dfrac{1}{2};\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\right]\)

\(\Rightarrow y_{min}=-\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\Leftrightarrow sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)=-1\Leftrightarrow...\)

\(y_{max}=\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\Leftrightarrow sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)=1\Leftrightarrow...\)

Không có max

`a)sqrt{x^2-2x+5}`

`=sqrt{x^2-2x+1+4}`

`=sqrt{(x-1)^2+4}`

Vì `(x-1)^2>=0`

`=>(x-1)^2+4>=4`

`=>sqrt{(x-1)^2+4}>=sqrt4=2`

Dấu "=" xảy ra khi `x=1.`

`b)2+sqrt{x^2-4x+5}`

`=2+sqrt{x^2-4x+4+1}`

`=2+sqrt{(x-2)^2+1}`

Vì `(x-2)^2>=0`

`=>(x-2)^2+1>=1`

`=>sqrt{(x-2)^2+1}>=1`

`=>sqrt{(x-2)^2+1}+2>=3`

Dấu "=" xảy ra khi `x=2`

\(2x^2-4xy+4y^2+2x+5=\left(x^2-4xy+4y^2\right)+\left(x^2+2x+1\right)+4=\left(x-2y\right)^2+\left(x+1\right)^2+4\)

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

\(\Rightarrow\left(x-2y\right)^2+\left(x+1\right)^2+4\ge4\)

vậy max của biểu thức trên = 4