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a) \(A=x^2-6x+11\)

\(\Rightarrow A=x^2-6x+9+2\)

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

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

\(\Rightarrow\left(x-3\right)^2+2\ge2\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\) x = 3

Vậy \(MIN\) \(A=2\Leftrightarrow x=3\)

b) \(B=2x^2+10x-1\)

\(\Rightarrow B=2\left(x^2+5\right)-1\)

\(\Rightarrow B=2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)-\dfrac{25}{2}-1\)

\(\Rightarrow B=2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)-\dfrac{23}{2}\)

Ta có: \(2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)\ge0\forall x\)

\(\Rightarrow2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)-\dfrac{23}{2}\ge-\dfrac{23}{2}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\) x = \(\dfrac{-5}{2}\)

Vậy \(MIN\) \(B=\dfrac{-23}{2}\Leftrightarrow x=\dfrac{-5}{2}\)

c) \(C=5x-x^2\)

\(\Rightarrow C=-\left(x^2-5x\right)\)

\(\Rightarrow C=-\left(x^2-2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)+\dfrac{25}{4}\)

\(\Rightarrow C=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\)

Ta có: \(-\left(x-\dfrac{5}{2}\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\) x = \(\dfrac{5}{2}\)

Vậy \(MAX\) \(C=\dfrac{25}{4}\Leftrightarrow x=\dfrac{5}{2}\)

16 tháng 12 2014

a) 2

b) 25/4

c)  -9/2

3 tháng 8 2020

a) \(A=x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\)

\(\left(x-3\right)^2\ge0\forall x\Rightarrow\left(x-3\right)^2+2\ge2\)

Đẳng thức xảy ra <=> x - 3 = 0 => x = 3

Vậy AMin = 2 , đạt được khi x = 3

b) \(B=5x-x^2=-x^2+5x=-x^2+5x-\frac{25}{4}+\frac{25}{4}=-\left(x^2-5x+\frac{25}{4}\right)+\frac{25}{4}=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\)

\(-\left(x-\frac{5}{2}\right)^2\le0\forall x\Rightarrow-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)

Đẳng thức xảy ra <=> x - 5/2 = 0 => x = 5/2

Vậy BMax = 25/4 , đạt được khi x = 5/2

c) \(2x-2x^2-5=-2x^2+2x-5=-2\left(x^2-x+\frac{1}{4}\right)-\frac{9}{2}=-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\)

\(-2\left(x-\frac{1}{2}\right)^2\le0\forall x\Rightarrow-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\le-\frac{9}{2}\)

Đẳng thức xảy ra <=> x - 1/2 = 0 => x = 1/2

Vậy CMax = -9/2 , đạt được khi x = 1/2

22 tháng 10 2021

\(1,a,A=x^2-6x+25\)

\(=x^2-2.x.3+9-9+25\)

\(=\left(x-3\right)^2+16\)

Ta có :

\(\left(x-3\right)^2\ge0\)Với mọi x

\(\Rightarrow\left(x-3\right)^2+16\ge16\)

Hay \(A\ge16\)

\(\Rightarrow A_{min}=16\)

\(\Leftrightarrow x=3\)

22 tháng 10 2021

\(b,B=4x^2+4x-2\)

\(B=4x^2+4x+1-3\)

\(B=\left(4x^2+4x+1\right)-3\)

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

Ta có : 

\(\left(2x+1\right)^2\ge0\)với mọi x

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

\(\Leftrightarrow B\ge-3\)

\(\Rightarrow B_{min}=-3\)

\(\Leftrightarrow x=-\frac{1}{2}\)

11 tháng 8 2016

\(a,A=x^2-6x+11=\left(x-3\right)^2+2\)\(\Leftrightarrow Amin=2\)

Dấu = xảy ra \(\Leftrightarrow x=3\)

11 tháng 8 2016

\(2x^2+10x-1=2\left(x^2+5x-\frac{1}{2}\right)=2\left(x^2+2.\frac{5}{2}x+\frac{25}{4}-\frac{27}{4}\right)=2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\)

\(\Rightarrow Bmin=\frac{-27}{2}.''=''\Leftrightarrow x=\frac{-5}{2}\)

17 tháng 10 2016

a)\(A=4x^2+4x+11\)

\(=4x^2+4x+1+10\)

\(=\left(2x+1\right)^2+10\ge10\)

Dấu = khi \(x=\frac{-1}{2}\)

Vậy MinA=10 khi \(x=\frac{-1}{2}\)

b)\(B=3x^2-6x+1\)

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

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

\(=3\left(x-1\right)^2-2\ge-2\)

Dấu = khi \(x=1\)

Vậy MinB=-2 khi \(x=1\)

c)\(C=x^2-2x+y^2-4y+6\)

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

\(=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)

Dấu = khi \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

Vậy MinC=1 khi \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

16 tháng 7 2017

a, Ta có: \(B=2x^2+10x-1=2x^2+10x+\dfrac{25}{2}-\dfrac{27}{2}\)

\(=2\left(x^2+2.x.\dfrac{5}{2}+\dfrac{25}{4}\right)-\dfrac{27}{2}\)

\(=2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}\ge\dfrac{-27}{2}\)

Dấu " = " khi \(2\left(x+\dfrac{5}{2}\right)^2=0\Leftrightarrow x=\dfrac{-5}{2}\)

Vậy \(MIN_B=\dfrac{-27}{2}\) khi \(x=\dfrac{-5}{2}\)

b, Ta có: \(C=5x-x^2=-\left(x^2-2.x.\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{25}{4}\right)\)

\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)

Dấu " = " khi \(-\left(x-\dfrac{5}{2}\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)

Vậy \(MAX_C=\dfrac{25}{4}\) khi \(x=\dfrac{5}{2}\)

8 tháng 3 2021

\(A=x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1\forall x\)

Dấu "=" xảy ra <=> x = 3

Vậy MinA = 1

\(B=5x^2-10x+3=5\left(x^2-2x+1\right)-2=5\left(x-1\right)^2-2\ge-2\forall x\)

Dấu "=" xảy ra <=> x = 1

Vậy MinB = -2

\(C=2x^2+8x+y^2-10y+43=2\left(x^2+4x+4\right)+\left(y^2-10y+25\right)+10=2\left(x+2\right)^2+\left(y-5\right)^2+10\ge10\forall x,y\)

Dấu "=" xảy ra <=> x = -2 ; y = 5

Vậy MinC = 10

8 tháng 3 2021

\(A=x^2-6x+10\)

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

\(=\left(x-3\right)^2+1\ge1\forall x\)

Dấu"=" xảy ra khi \(x-3=0\Leftrightarrow x=3\)

Vậy \(Min_A=1\Leftrightarrow x=3\)

b,\(B=5x^2-10x+3\)

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

\(=5\left(x-1\right)^2-2\ge-2\forall x\)

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

Vậy \(Min_B=-2\Leftrightarrow x=1\)

c,\(C=2x^3+8x+y^2-10+43\)

\(=2x^2+8x+8+y^2-10y+25+10\)

\(=2\left(x^2+4x+4\right)+\left(y^2-10y+25\right)+10\)

\(=2\left(x+2\right)^2+\left(y-5\right)^2+10\ge10\forall x,y\)

Dấu"=" xảy ra khi \(\orbr{\begin{cases}x+2=0\\y-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\y=5\end{cases}}}\)

Vậy \(Min_C=10\Leftrightarrow x=-2;y=5\)

18 tháng 9 2017

Ta có : A = x2 - 6x + 15 

= x2 - 6x + 9 + 6

= (x - 3)2 + 6 \(\ge6\forall x\in R\)

Vậy Amin = 6 khi x = 3.