`A=-x^2+2x+10`

`=-(x^2-2x)+10`

`=-(x-1)^2+11<=11`

Dấu "=" xảy ra khi `x=1`.

`B=4x-2x^2+8`

`=-2(x^2-2x)+8`

`=-2(x^2-2x+1)+10`

`=-2(x-1)^2+10<=10`

Dấu "=" xảy ra khi `x=1`

`C=-x^2-x+1`

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

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

`=-(x+1/2)^2+5/4<=5/4`

Dấu "=" xảy ra khi `x=-1/2`

`D=-4x^2+6x+3`

`=-(4x^2-6x)+3`

`=-(4x^2-6x+9/4)+21/4`

`=-(2x-3/2)^2+21/4<=21/4`

Dấu "=' xảy ra khi `2x=3/2<=>x=3/4`

\(a,A=-x^2+2x+10=-x^2+2x-1+11=-\left(x^2-2x+1\right)+11\)

\(=11-\left(x-1\right)^2\)

- Thấy : \(\left(x-1\right)^2\ge0\forall x\in R\)

\(\Rightarrow A=11-\left(x-1\right)^2\le11\)

Vậy MaxA = 11 <=> x = 1 .

\(b,B=-2x^2+4x-2+10=-2\left(x^2-2x+1\right)+10=10-2\left(x-1\right)^2\)

- Thấy : \(\left(x-1\right)^2\ge0\forall x\in R\)

\(\Rightarrow B=10-2\left(x-1\right)^2\le10\)

Vậy MaxB = 10 <=> x = 1 .

\(c,C=-x^2-\dfrac{1}{2}.2.x-\dfrac{1}{4}+\dfrac{5}{4}=\dfrac{5}{4}-\left(x+\dfrac{1}{2}\right)^2\)

- Thấy : \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\in R\)

\(\Rightarrow C=\dfrac{5}{4}-\left(x+\dfrac{1}{2}\right)^2\le\dfrac{5}{4}\)

Vậy MaxC = 5/4 <=> x = -1/2 .

\(d,D=-4x^2+6x+3=-4x^2+2x.2.\dfrac{6}{4}-\dfrac{9}{4}+\dfrac{21}{4}=-\left(4x^2-6x+\dfrac{9}{4}\right)+\dfrac{21}{4}\)

\(=\dfrac{21}{4}-\left(2x-\dfrac{3}{2}\right)^2\)

- Thấy : \(\left(2x-\dfrac{3}{2}\right)^2\ge0\forall x\in R\)

\(\Rightarrow A=\dfrac{21}{4}-\left(2x-\dfrac{3}{2}\right)^2\le\dfrac{21}{4}\)

Vậy MaxD=21/4 <=> x = 3/4 .

\(x^2+2x+3\)

\(=\left(x^2+2x+1\right)+2\)

\(=\left(x+1\right)^2+2\)

Do \(\left(x+1\right)^2\ge0\) với mọi x

\(\Rightarrow x^2+2x+3\ge2\)

Dấu = khi x=-1

A=x²-4x+1

=x²-4x+4-3

=(x-2)²-3

=> GTNN là : -3

B=đề sai.

C=chả hiểu gì hết.

1.

Đặt \(x-2=t\ne0\Rightarrow x=t+2\)

\(B=\dfrac{4\left(t+2\right)^2-6\left(t+2\right)+1}{t^2}=\dfrac{4t^2+10t+5}{t^2}=\dfrac{5}{t^2}+\dfrac{2}{t}+4=5\left(\dfrac{1}{t}+\dfrac{1}{5}\right)^2+\dfrac{19}{5}\ge\dfrac{19}{5}\)

\(B_{min}=\dfrac{19}{5}\) khi \(t=-5\) hay \(x=-3\)

2.

Đặt \(x-1=t\ne0\Rightarrow x=t+1\)

\(C=\dfrac{\left(t+1\right)^2+4\left(t+1\right)-14}{t^2}=\dfrac{t^2+6t-9}{t^2}=-\dfrac{9}{t^2}+\dfrac{6}{t}+1=-\left(\dfrac{3}{t}-1\right)^2+2\le2\)

\(C_{max}=2\) khi \(t=3\) hay \(x=4\)

`A=(2x)^2+2.2x.1+1^2+1=(2x+1)^2+1`

`=> A_(min)=1 <=>x=-1/2`

`B=(\sqrt2x)^2-2.\sqrt2 x . \sqrt2/2 + (\sqrt2/2)^2 + 1/2`

`=(\sqrt2x-\sqrt2/2)^2+1/2`

`=> B_(min)=1/2 <=> x=1/2`

`C=-(x^2-2.x.3+3^2+6)=-(x-3)^2-6`

`=> C_(max)=-6 <=> x=3`