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

Dấu = xảy ra \(\Leftrightarrow x=3\)

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

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

\(5x-x^2=-\left(x^2-5x\right)=-\left(x^2-2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}\right)=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\)\(\Leftrightarrow Cmax=\frac{25}{4}.''=''\Leftrightarrow x=\frac{5}{2}\)

Mình chỉ tìm giá trị chứ không tìm x đâu nhé (đề bài ghi thế)

a) 

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

\(\left(x-3\right)^2\ge0\forall x\\ 2\ge2\\ \Rightarrow\left(x-3\right)^2+2\ge2\forall x\\ A\ge2\forall x\\ \Rightarrow A_{min}=2\)

b) B = 2x² + 10 - 1

B = 2(x² + 5) - 1

B = 2(x² + 2.\(\frac{5}{2}\).x + \(\frac{25}{4}\)) -  \(\frac{25}{2}\) - 1

B = 2(x + \(\frac{5}{2}\))² - \(\frac{27}{2}\)

Vậy GTNN của B = \(\frac{-27}{2}\) khi x = \(\frac{-5}{2}\).

c) C = 5x - x²

C = -(x² - 5x)

C = -(x² - 2.\(\frac{5}{2}\).x + \(\frac{25}{4}\)) + \(\frac{25}{4}\)

C = -(x - \(\frac{5}{2}\))² + \(\frac{25}{4}\)

Vậy GTLN của C = \(\frac{25}{4}\) khi x = \(\frac{5}{2}\).

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

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

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

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

Vậy GTNN của \(A=2\) khi \(x-3=0\Leftrightarrow x=3\)

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

\(\Leftrightarrow B=2x^2+10x+\dfrac{25}{2}-\dfrac{27}{2}\)

\(\Leftrightarrow B=\left(2x^2+10x+\dfrac{25}{2}\right)-\dfrac{27}{2}\)

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

\(\Leftrightarrow B=2\left[x^2+2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]-\dfrac{27}{2}\)

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

Vậy GTNN của \(B=\dfrac{-27}{2}\) khi \(x+\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{-5}{2}\)

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

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

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

\(\Leftrightarrow C=-\left[x^2-2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]+\dfrac{25}{4}\)

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

Vậy GTLN của \(C=\dfrac{25}{4}\) khi \(x-\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{5}{2}\)

A=11

B=9

C=0

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

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

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

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

Dấu = xảy ra khi: \(\left(x-3\right)^2=0\Rightarrow x-3=0\Rightarrow x=3\)

Vậy: \(Min_A=2\) tại \(x=3\)

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

\(B=2x^2+10x+\frac{25}{2}-\frac{27}{2}\)

\(B=\left(\sqrt{2}x-\sqrt{\frac{25}{2}}\right)^2-\frac{27}{2}\)

Có: \(\left(\sqrt{2}x-\sqrt{\frac{25}{2}}\right)^2\ge0\Rightarrow\left(\sqrt{2}x-\sqrt{\frac{25}{2}}\right)^2-\frac{27}{2}\ge-\frac{27}{2}\)

Dấu = xảy ra khi: \(\left(\sqrt{2}x-\sqrt{\frac{25}{2}}\right)^2=0\Rightarrow\sqrt{2}x-\sqrt{\frac{25}{2}}=0\Rightarrow x=\frac{5}{2}\)

Vậy: \(Min_B=-\frac{27}{2}\) tại \(x=\frac{5}{2}\)

c) \(C=5x-x^2\) 

\(C=\frac{25}{4}-x^2+5x-\frac{25}{4}\)

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

Có: \(\left(x-\frac{5}{2}\right)^2\ge0\Rightarrow\frac{25}{4}-\left(x-\frac{5}{2}\right)^2\le\frac{25}{4}\)

Dấu = xảy ra khi: \(\left(x-\frac{5}{2}\right)^2=0\Rightarrow x-\frac{5}{2}=0\Rightarrow x=\frac{5}{2}\)

Vậy: \(Max_C=\frac{25}{4}\) tại \(x=\frac{5}{2}\)

a/ \(A=x^2-6x+11=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\)

b/ \(B=2x^2+10-1=2x^2+9\ge9\)

c/ \(C=5x-x^2=\left(-x^2+\frac{2\times5x}{2}-\frac{25}{4}\right)+\frac{25}{4}\)

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

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

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 - 6x + 11 

Nhập phương trình vào máy tính lặp 3 lần  dấu =

GTNN của A = 3

B = 2x² + 10x - 1

Nhập phương trình vào máy tính lặp 3 lần dấu =

GTNN của B = \(-\frac{5}{2}\)

C = 5x - x² 

=> C = -x² + 5x

Nhập phương trình vào máy tính lặp 3 lần dấu =

GTLN của C = \(\frac{5}{2}\)

Trả lời

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hok tốt

a) 2

b) 25/4

c)  -9/2

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 A_Min = 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 B_Max = 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 C_Max = -9/2 , đạt được khi x = 1/2

a: Ta có: \(x^2+x+1\)

\(=x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)

b: Ta có: \(-x^2+x+2\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{9}{4}\right)\)

\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)