Ta có: \(x^2-6x+11\)

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

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

Dấu '=' xảy ra khi x=3

Ta có:

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

Áp dụng bđt Minkowski, ta có:

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

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

\(A=\sqrt{4^2+\left(\sqrt{2}+\sqrt{3}\right)^2\left(y+1\right)^2}\ge\sqrt{4^2}=4\)

\(\Rightarrow A\ge4.Đ\text{TXR}\Leftrightarrow\orbr{\begin{cases}x=1;y=-1\\x=3;y=-1\end{cases}}\)

Dấu "=" xảy ra khi (x; y) = (3; -1)

Tìm min và max cả 2 câu hả bạn

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

Vậy GTNN của A là 2 khi x = 3

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

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

Xét \(A\ge-\frac{1}{2}\)

<=> \(\frac{6x+11}{x^2-2x+3}\ge-\frac{1}{2}\)

<=> \(x^2-2x+3\ge-12x-22\)

<=> \(x^2+10x+25\ge0\)<=> \(\left(x+5\right)^2\ge0\)(luôn đúng) 

Vậy \(MinA=-\frac{1}{2}\)khi x=-5

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

\(A=x^2-6x+9+x^2-22x+121\)

\(A=2x^2-28x+130\)

\(A=2\left(x^2-14x+49\right)+32\)

\(A=2\left(x-7\right)^2+32\ge32\)

Vậy GTNN của A là 32 khi x = 7

\(A=19-6x-9x^2 \)

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

\(A=-\left(3x+1\right)^2+20\le20\)

Vậy GTLN của A là 20 khi x = \(-\frac{1}{3}\)

a) \(3x^2+2y⋮11\Leftrightarrow16\left(3x^2+2y\right)⋮11\Leftrightarrow48x^2-33x^2+32y-44y⋮11\)

\(\Leftrightarrow15x^2-12y⋮11\)

b) \(2x+3y^2⋮7\Leftrightarrow10\left(2x+3y^2\right)⋮7\Leftrightarrow20x-14x+30y^2-14y^2⋮7\)

\(\Leftrightarrow6x+16y^2⋮7\)

1) A = 3 - 4x² - 4x  = - (4x² + 4x +1) + 4 = - (2x+1)² + 4 

Vì  - (2x+1)² \(\le\)0 nên A =  - (2x+1)² + 4 \(\le\) 4 vậy maxA = 4 khi 2x+1 = 0 => x = -1/2

b) ta có x² + 6x + 11 = x² + 2.3x + 9 + 2 = (x+3)² + 2 \(\ge\) 0 + 4 = 4

=> \(B=\frac{1}{x^2+6x+11}\le\frac{1}{4}\) vậy maxB = 1/4 khi x = -3

2) a) 3x² - 3x + 1 = 3.(x² - x) + 1 = 3.(x² - 2.x\(\frac{1}{2}\) + \(\frac{1}{4}\)) + \(\frac{1}{4}\) = 3.(x - \(\frac{1}{2}\) )² + \(\frac{1}{4}\) \(\ge\)0 + \(\frac{1}{4}\)= \(\frac{1}{4}\)

vậy min(3x² - 3x + 1) = 1/4 khi x = 1/2

b) Áp dụng bất đẳng thức giá trị tuyệt đối: |a| + |b| \(\ge\) |a - b|. dấu = khi a.b < 0

ta có:  |3x - 3| + |3x - 5| \(\ge\) |3x - 3 - (3x - 5)| = |2| = 2

vậy min = 2 khi (3x - 3)(3x - 5) < 0 hay 1< x <  5/3

\(A=4x^2-12x+11\)

\(A=\left(2x\right)^2-2.2x.3+3^2+2\)

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

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

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

Dấu = xảy ra \(\Leftrightarrow\left(2x-3\right)^2=0\Leftrightarrow2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Vậy A_min=2\(\Leftrightarrow x=\frac{3}{2}\)

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

\(B=\left(x^2-2x+1\right)+\left(y^2+2.2y+2^2\right)+1\)

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

Ta có:  \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{cases}\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\forall x;y}\)

Dấu = xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=-2\end{cases}}}\)

Vậy B_min=1\(\Leftrightarrow x=1;y=-2\)

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

\(\Rightarrow-A=x^2+6x-1\)

\(-A=\left(x^2+2.3x+3^2\right)-10\)

\(-A=\left(x+3\right)^2-10\)

\(\Rightarrow A=-\left(x+3\right)^2+10\)

Ta có: \(\left(x+3\right)^2\ge0\forall x\Rightarrow-\left(x+3\right)^2\le0\forall x\Rightarrow-\left(x+3\right)^2+10\le10\forall x\)

Dấu = xảy ra \(\Leftrightarrow-\left(x+3\right)^2=0\Leftrightarrow\left(x+3\right)^2=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

Vậy A_max=10\(\Leftrightarrow\)x= -3

Sửa đề:

\(B=-2x^2-8x-6\)

\(B=-2.\left(x^2+2.2x+2^2\right)+2\)

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

Ta có: \(2.\left(x+2\right)^2\ge0\forall x\Rightarrow-2.\left(x+2\right)^2\le0\forall x\Rightarrow-2.\left(x+2\right)^2+2\le2\forall x\)

Dấu = xảy ra \(\Leftrightarrow-2.\left(x+2\right)^2=0\Leftrightarrow\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

Vậy B_max=2\(\Leftrightarrow x=-2\)

Đề phải là tìm min mới đúng

a, A=4x²-12x+11

=(4x²-12x+9)+2

=(2x-3)²+2

Vì (2x-3)² \(\ge\) 0 => A=(2x-3)²+2 \(\ge\) 2

Dấu "=" xảy ra khi 2x-3=0 <=> x=3/2

Vậy Amin = 2 khi x=3/2

b, B=x²-2x+y²+4y+6

=(x²-2x+1)+(y²+4y+4)+1

=(x-1)²+(y+2)²+1

Vì \(\left(x-1\right)^2\ge0;\left(y+2\right)^2\ge0\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\)

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

Dấu "=" xảy ra khi x=1,y=-2

Vậy Bmin = 1 khi x=1,y=-2