`A=sqrt{x-2}+sqrt{6-x}(2<=x<=6)`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>A>=sqrt{x-2+6-x}=2`
Dấu "=" `<=>x=2` hoặc `x=6`
Áp dụng BĐT bunhia
`=>A<=sqrt{2(x-2+6-x)}=2sqrt2`
Dấu "=" `<=>x=4`
`C=sqrt{1+x}+sqrt{8-x}(-1<=x<=8)`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>A>=sqrt{1+x+8-x}=3`
Dấu "=" `<=>x=-1` hoặc `x=8`
Áp dụng BĐT bunhia
`=>A<=sqrt{2(1+x+8-x)}=3sqrt2`
Dấu "=" `<=>x=7/2`

`D=2sqrt{x+5}+sqrt{1-2x}(-5<=x<=1/2)`
`=sqrt{4x+20}+sqrt{1-2x}`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>D>=sqrt{4x+20+1-2x}=sqrt{2x+21}`
Mà `x>=-5`
`=>D>=sqrt{-10+21}=sqrt{11}`
Dấu "=" `<=>x=-5`

\(A=\left(x-1\right)^2+8\ge8\\ A_{min}=8\Leftrightarrow x=1\\ B=\left(x+3\right)^2-12\ge-12\\ B_{min}=-12\Leftrightarrow x=-3\\ C=x^2-4x+3+9=\left(x-2\right)^2+8\ge8\\ C_{min}=8\Leftrightarrow x=2\\ E=-\left(x+2\right)^2+11\le11\\ E_{max}=11\Leftrightarrow x=-2\\ F=9-4x^2\le9\\ F_{max}=9\Leftrightarrow x=0\)

\(a,\\ A=25x^2-10x+11\\ =\left(5x\right)^2-2.5x.1+1^2+10\\ =\left(5x+1\right)^2+10\ge10\forall x\in R\\ Vậy:min_A=10.khi.5x+1=0\Leftrightarrow x=-\dfrac{1}{5}\\ B=\left(x-3\right)^2+\left(11-x\right)^2\\ =\left(x^2-6x+9\right)+\left(121-22x+x^2\right)\\ =x^2+x^2-6x-22x+9+121=2x^2-28x+130\\ =2\left(x^2-14x+49\right)+32\\ =2\left(x-7\right)^2+32\\ Vì:2\left(x-7\right)^2\ge0\forall x\in R\\ Nên:2\left(x-7\right)^2+32\ge32\forall x\in R\\ Vậy:min_B=32.khi.\left(x-7\right)=0\Leftrightarrow x=7\\Tương.tự.cho.biểu.thức.C\)

b:

\(D=-25x^2+10x-1-10\)

\(=-\left(25x^2-10x+1\right)-10\)

\(=-\left(5x-1\right)^2-10< =-10\)

Dấu = xảy ra khi x=1/5

\(E=-9x^2-6x-1+20\)

\(=-\left(9x^2+6x+1\right)+20\)

\(=-\left(3x+1\right)^2+20< =20\)

Dấu = xảy ra khi x=-1/3

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

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

Dấu = xảy ra khi x=1

A = x² - 2x + 9 = ( x² - 2x + 1 ) + 8 = ( x - 1 )² + 8 ≥ 8 ∀ x

Dấu "=" xảy ra khi x = 1

=> MinA = 8 <=> x = 1

B = x² + 6x - 3 = ( x² + 6x + 9 ) - 12 = ( x + 3 )² - 12 ≥ -12 ∀ x

Dấu "=" xảy ra khi x = -3

=> MinB = -12 <=> x = -3

C = ( x - 1 )( x - 3 ) + 9 = x² - 4x + 3 + 9 = ( x² - 4x + 4 ) + 8 = ( x - 2 )² + 8 ≥ 8 ∀ x

Dấu "=" xảy ra khi x = 2

=> MinC = 8 <=> x = 2

D = -x² - 4x + 7 = -( x² + 4x + 4 ) + 11 = -( x + 2 )² + 11 ≤ 11 ∀ x

Dấu "=" xảy ra khi x = -2

=> MaxD = 11 <=> x = -2

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a,\(A=2\sqrt{x^2+x+\dfrac{1}{2}}=2\sqrt{x^2+x+\dfrac{1}{4}+\dfrac{1}{4}}=2\sqrt{\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{4}}\)

\(=\sqrt{4\left(x+\dfrac{1}{2}\right)^2+1}\ge1\) dấu"=" xảy ra<=>x=-1/2

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

\(=\sqrt{2\left(x-1\right)^2+3}\ge\sqrt{3}\) dấu"=" xảy ra<=>x=1

\(C=\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\ge\dfrac{-2}{-\sqrt{2}}=\sqrt{2}\) dấu"=" xảy ra<=>x=1

\(D=x-2\sqrt{x+2}\ge-2\) dấu"=" xảy ra<=>x=-2

d)D=\(x-2\sqrt{x+2}=\left(x+2\right)-2\sqrt{x+2}+1-3\)

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

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x+2}=1\Leftrightarrow x+2=1\Leftrightarrow x=-1\)

A = | x - 3 | + 1

Ta có : \(\left|x-3\right|\ge0\forall x\Rightarrow\left|x+3\right|+1\ge1\)

Dấu = xảy ra <=> | x + 3 | = 0

                      <=> x + 3 = 0

                      <=> x = -3

Vậy A_Min = 1 khi x = -3

B = -100 - | 7 - x |

Ta có : \(\left|7-x\right|\ge0\forall x\Rightarrow-\left|7-x\right|\le0\)

=> \(-100-\left|7-x\right|\le-100\)

Dấu = xảy ra <=> - | 7 - x | = 0

                     <=> 7 - x = 0

                     <=> x = 7

Vậy B_Max = -100 khi x = 7

C = -( x + 1 )² - | 2 - y | + 11

Ta có : \(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\\left|2-y\right|\ge0\forall y\end{cases}\Rightarrow}\hept{\begin{cases}-\left(x+1\right)^2\le0\\-\left|2-y\right|\le0\end{cases}}\)

=> \(-\left(x+1\right)^2-\left|2-y\right|\le11\forall x,y\)

Dấu = xảy ra <=> -( x + 1 )² = 0 và | 2 - y | = 0

                     <=> x + 1 = 0 và 2 - y = 0

                     <=> x = -1 và y = 2

Vậy C_Max = 11 khi x = -1 ; y = 2

D = ( x - 1 )² + | 2y + 2 | + 3

Ta có : \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left|2y+2\right|\ge0\forall y\end{cases}\Rightarrow\left(x-1\right)^2+\left|2y+2\right|+3\ge}3\)

Dấu = xảy ra <=> ( x - 1 )² = 0 và | 2y + 2 | = 0

                      <=> x - 1 = 0 và 2y + 2 = 0

                      <=> x = 1 và y = -1

Vậy D_Min = 3 khi x = 1 và y = -1

a) A=/x-3/+1>=0+1=1

dấu "="sảy ra <=>x-3=0<=>x=3

vậy min A=1 <=>x=3

b) B=-100-/7-x/=<-100-0=-100

dấu "="sảy ra <=>7-x=0<=>x=7

vậy max B=-100<=>x=7

c)C=-(x+1)^2-/2-y/+11=<-0-0+11=11

dấu "="sảy ra <=>x=-1vày=2

vậy max C=11<=>x=-1 và y=-2

d)D=(x-1)^2+/2y+2/+3>=0+0+3=3

dấu "="sảy ra <=>x=1 và y =-1

vậy min D=3<=>x=1 và y=-1

Sửa đề:

A=/x+5/+10

Ta có: /x+5/>= 0 với mọi x>=0

=> A=/x+5/+10 >= 10

=> Amin=10. Dấu "=" xảy ra <=> x+5=0<=> x=-5

Vậy...

\(\text{a) }A=\left|x+5\right|+10\)

\(\text{Vì }\left|x+5\right|\ge0\forall x\)

\(\Rightarrow A=\left|x+5\right|+10\ge10\)

\(\text{Dấu ''='' xảy ra khi :}\)

\(\left|x+5\right|=0\)

\(\Rightarrow x=-5\)

\(\text{Vậy Min}_A=10\Leftrightarrow x=-5\)

\(\text{b) }\left|3-x\right|+5\)

\(\text{Vì }\left|3-x\right|\ge0\forall x\)

\(\Rightarrow\left|3-x\right|+5\ge5\)

\(\text{Dấu ''='' xảy ra khi :}\)

\(\left|3-x\right|=0\)

\(\Rightarrow x=3\)

\(\text{Vậy Min}_B=5\Leftrightarrow x=3\)

\(\text{d) }D=\left(x+2\right)^2+15\)

\(\text{Vì ( x + 2 )}^2\ge0\forall x\)

\(\Rightarrow\left(x+2\right)^2+15\ge15\)

\(\text{Dấu ''='' xảy ra khi :}\)

\(\left(x+2\right)^2=0\)

\(\Rightarrow x+2=0\)

\(\Rightarrow x=-2\)

\(C=\left|x+1\right|+\left|y+4\right|\)

Ta có +) \(\left|x+1\right|\ge0\).Dấu "=" xảy ra \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

+)\(\left|y+4\right|\ge0\).Dấu "=" xảy ra \(\Leftrightarrow y+4=0\Leftrightarrow y=-4\)

\(\Rightarrow B=\left|x+1\right|+\left|y+2\right|\ge0\)

\(\Rightarrow minB=0\Leftrightarrow\hept{\begin{cases}x=-1\\y=-4\end{cases}}\)

P/s: ko chắc nha