a) \(A=\sqrt[]{x^2-2x+5}\)

\(\Leftrightarrow A=\sqrt[]{x^2-2x+1+4}\)

\(\Leftrightarrow A=\sqrt[]{\left(x+1\right)^2+4}\)

mà \(\left(x+1\right)^2\ge0,\forall x\in R\)

\(A=\sqrt[]{\left(x+1\right)^2+4}\ge\sqrt[]{4}=2\)

Dấu "=" xảy ra khi và chỉ khi \(x+1=0\Leftrightarrow x=-1\)

Vậy \(GTNN\left(A\right)=2\left(khi.x=-1\right)\)

b) \(B=5-\sqrt[]{x^2-6x+14}\)

\(\Leftrightarrow B=5-\sqrt[]{x^2-6x+9+5}\)

\(\Leftrightarrow B=5-\sqrt[]{\left(x-3\right)^2+5}\left(1\right)\)

Ta có : \(\left(x-3\right)^2\ge0,\forall x\in R\)

\(\Leftrightarrow\left(x-3\right)^2+5\ge5,\forall x\in R\)

\(\Leftrightarrow\sqrt[]{\left(x-3\right)^2+5}\ge\sqrt[]{5},\forall x\in R\)

\(\Leftrightarrow-\sqrt[]{\left(x-3\right)^2+5}\le-\sqrt[]{5},\forall x\in R\)

\(\Leftrightarrow B=5-\sqrt[]{\left(x-3\right)^2+5}\le5-\sqrt[]{5},\forall x\in R\)

Dấu "=" xả ra khi và chỉ khi \(x-3=0\Leftrightarrow x=3\)

Vậy \(GTLN\left(B\right)=5-\sqrt[]{5}\left(khi.x=3\right)\)

Biểu thức nào em?

\(Q=-2\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{2}\le\dfrac{25}{2}\)

\(Q_{max}=\dfrac{25}{2}\) khi \(x=\dfrac{3}{2}\)

\(A=\dfrac{9\left(x^2+2\right)-9x^2+6x-1}{x^2+2}=9-\dfrac{\left(3x-1\right)^2}{x^2+2}\le9\)

\(A_{max}=9\) khi \(x=\dfrac{1}{3}\)

\(A=\dfrac{12x+34}{2\left(x^2+2\right)}=\dfrac{-\left(x^2+2\right)+x^2+12x+36}{2\left(x^2+2\right)}=-\dfrac{1}{2}+\dfrac{\left(x+6\right)^2}{2\left(x^2+2\right)}\le-\dfrac{1}{2}\)

\(A_{min}=-\dfrac{1}{2}\) khi \(x=-6\)

`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`

a, x^2-2x+1+4=(x-1)^2+4>=4. dấu = xảy ra khi x=1

b,dưa 2 ra làm tt

c, đưa dấu - ra

d nhân ra là đc

Bạn học hằng đẳng thức chưa?

\(x^2-2x+5\)

\(=x^2-2x+1+4\)

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

Ta có: \(\left(x-1\right)^2\ge0\forall x\)

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

\(x^2-2x+5=4\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy Min \(x^2-2x+5=4\Leftrightarrow x=1\)

Câu b trình bày tương tự, mình chỉ gợi ý

\(2x^2+6x-5=2\left(x^2+2.x.1,5+1,5^2\right)-9,5=2\left(x+1,5\right)^2-9,5\)

a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)

\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)

\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)

Vậy Max_Q=10 khi x=2, y=-2

b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)

\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)

Vậy Max_A=14 khi x=-3

+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)

\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)

\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)

Vậy Max_B=5 khi x=-1/2, y=1/3

c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)

\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)

Vậy Min_P=2 khi x=1, y=-3

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

\(\Leftrightarrow A=-\left(x^2-x-1\right)\)

\(\Leftrightarrow A=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}-\frac{5}{4}\right)\)

\(\Leftrightarrow-A=\left[\left(x-\frac{1}{2}\right)^2-\frac{5}{4}\right]\)

Ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\left(x-\frac{1}{2}\right)^2-\frac{5}{4}\ge\frac{-5}{4}\)hay \(-A\ge\frac{-5}{4}\)

\(\Rightarrow A\le\frac{5}{4}\)

Vậy \(A_{max}=\frac{5}{4}\)(Dấu "="\(\Leftrightarrow x=\frac{1}{2}\))

\(D=4x^2+6x+1\)

\(D=\left(2x\right)^2+2.2x.\frac{3}{2}+\frac{9}{4}+1-\frac{9}{4}\)

\(D=\left(2x+\frac{9}{4}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)

Dấu = xảy ra khi : 

  \(2x+\frac{9}{4}=0\Rightarrow x=-\frac{9}{8}\)

Vậy D_min = - 5/ 4 tại x = -9/8

\(C=-3\left(x^2-2x+1\right)+5\)

    \(=-3\left(x-1\right)^2+5\le5\forall x\)

Dấu"=" xảy ra<=> \(-3\left(x-1\right)^2=0\Leftrightarrow x=1\) 

Vậy.........

\(B=2\left(x^2-\frac{7}{2}x+\frac{49}{16}\right)-\frac{57}{8}\) 

   \(=2\left(x-\frac{7}{4}\right)^2-\frac{57}{8}\ge\frac{-57}{8}\forall x\) 

Dấu"="xảy ra<=> \(2\left(x-\frac{7}{4}\right)^2=0\Leftrightarrow x=\frac{7}{4}\) 

Vậy...

Lời giải:
$C=-15-x^2+6x=-6-(x^2-6x+9)=-6-(x-3)^2$

Vì $(x-2)^2\geq 0$ với mọi $x\in\mathbb{R}$

$\Rightarrow C\leq -6< 0$

Vậy $C$ luôn âm.