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)

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

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

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

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

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

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 = \(-\frac{1}{3}\)

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

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

\(=x\left(x-6\right)+11\)

\(=>A_{Min}=11\Leftrightarrow\hept{\begin{cases}x=0\\x=6\end{cases}}\)

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

\(C=\left(9x^2-6x+1\right)+4=\left(3x-1\right)^2+4\ge4\)

\(C_{min}=4\) khi \(x=\dfrac{1}{3}\)

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

\(D_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)

\(C=9x^2+5-6x=\left(9x^2-6x+1\right)+4=\left(3x-1\right)^2+4\ge4\)

\(minC=4\Leftrightarrow x=\dfrac{1}{3}\)

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

\(minD=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)

Ta có : 

\(A=\dfrac{6x+8}{x^2+1}\)

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

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

Vì \(\left(x+3\right)^2\ge0\) nên \(\dfrac{\left(x+3\right)^2}{x^2+1}\)

nên \(\dfrac{\left(x+3\right)^2}{x^2+1}-1\ge-1\) hay \(A>-1\)

Dấu ' = ' xảy ra khi \(x=-3\)

Vậy \(A_{min}=-1\) khi \(x=-3\)

Ta có :

\(A=\dfrac{6x+8}{x^2+1}\)

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

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

Vì \(-\dfrac{\left(3x-1\right)^2}{x^2+1}\le0\) nên \(-\dfrac{\left(3x-1\right)^2}{x^2+1}+9\le9\)

Dấu '' = '' xảy ra khi \(x=\dfrac{1}{3}\)

Vậy \(A_{max}=9\) khi \(x=\dfrac{1}{3}\)

tham khảo