Bài 5:

a) \(A=x^2-4x+9=\left(x^2-4x+4\right)+5=\left(x-2\right)^2+5\ge5\)

\(minA=5\Leftrightarrow x=2\)

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

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

c) \(C=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

\(minC=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)

Bài 4:

a) \(M=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

\(maxM=7\Leftrightarrow x=2\)

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

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

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

\(maxP=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)

a: \(A=-4x^2+4x-1\)

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

\(=-\left(2x-1\right)^2\le0\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

b: \(B=-x^2+5x\)

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

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

Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)

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

\(=-\left(2x-1\right)^2\le0\)

\(maxA=0\Leftrightarrow x=\dfrac{1}{2}\)

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

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

\(maxB=\dfrac{25}{4}\Leftrightarrow x=\dfrac{5}{2}\)

c) \(C=-3x^2-9x+6=-3\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{51}{4}\)

\(=-3\left(x+\dfrac{3}{2}\right)^2+\dfrac{51}{4}\le\dfrac{51}{4}\)

\(maxC=\dfrac{51}{4}\Leftrightarrow x=-\dfrac{3}{2}\)

\(A=\dfrac{4x+3}{x^2+1}\Leftrightarrow Ax^2+A=4x+3\\ \Leftrightarrow Ax^2-4x+A-3=0\)

Coi đây là PT bậc 2 ẩn x thì PT có nghiệm

\(\Leftrightarrow\Delta=16-4A\left(A-3\right)\ge0\\ \Leftrightarrow16-4A^2+12A\ge0\\ \Leftrightarrow-A^2+3A+4\ge0\\ \Leftrightarrow-1\le A\le4\)

Vậy \(A_{max}=4;A_{min}=-1\)

\(A_{max}=4\Leftrightarrow\dfrac{4x+3}{x^2+1}=4\Leftrightarrow4x^2-4x+1=0\\ \Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\\ A_{min}=-1\Leftrightarrow\dfrac{4x+3}{x^2+1}=-1\Leftrightarrow x^2+1=-4x-3\Leftrightarrow x^2+4x+4=0\\ \Leftrightarrow\left(x+2\right)^2=0\Leftrightarrow x=-2\)

a,A=(2x)²-2.2x.2+2²+11=(2x-2)²+11

Vì (2x-2)²luôn lớn hơn hoặc bằng 0

=>A>hoặc =0+11 hay a>hoặc =11

vậy GTNN của A là 11 khi x=1

a: =x^2-6x+9+x^2-6x+9

=2(x-3)^2>=0

Dấu = xảy ra khi x=3

b: =-(x^2+4x+y^2-2y)

=-(x^2+4x+4+y^2-2y+1-5)

=-(x+2)^2-(y-1)^2+5<=5

Dấu = xảy ra khi x=-2 và y=1

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

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

vì -(x+2)^2 <=0

nên MIN A=0

<=>-(x+2)=0=>x=-2

vây min của A là 0 tại x=-2

A = 2x - x² - 4

A = - [ x² - 2 . 1 / 2 . x + ( 1 / 2 )² - ( 1 / 2 )² -  4 ]

A = - ( x - 1 / 2 )² - 17 / 4 \(\le\)- 17 / 4

Dấu = xảy ra \(\Leftrightarrow\)x - 1 / 2 = 0

                       \(\Rightarrow\)x = 1 / 2

Vậy : Min A = - 17 / 4 \(\Leftrightarrow\)x = 1 / 2

B với C nữa

Bài 1: 

a) Ta có: \(A=-x^2-4x-2\)

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

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

\(=-\left(x+2\right)^2+2\le2\forall x\)

Dấu '=' xảy ra khi x=-2

b) Ta có: \(B=-2x^2-3x+5\)

\(=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)

\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)

c) Ta có: \(C=\left(2-x\right)\left(x+4\right)\)

\(=2x+8-x^2-4x\)

\(=-x^2-2x+8\)

\(=-\left(x^2+2x-8\right)\)

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

\(=-\left(x+1\right)^2+9\le9\forall x\)

Dấu '=' xảy ra khi x=-1

Bài 2: 
a) Ta có: \(=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)

b) Ta có: \(B=9x^2-6xy+2y^2+1\)

\(=9x^2-6xy+y^2+y^2+1\)

\(=\left(3x-y\right)^2+y^2+1>0\forall x,y\)

c) Ta có: \(E=x^2-2x+y^2-4y+6\)

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

\(=\left(x-1\right)^2+\left(y-2\right)^2+1>0\forall x,y\)

a) Ta thấy: \(\left|\dfrac{2}{5}-x\right|\ge0\forall x\)

\(\Rightarrow Q=\dfrac{9}{2}+\left|\dfrac{2}{5}-x\right|\ge\dfrac{9}{2}\forall x\)

Dấu \("="\) xảy ra khi: \(\left|\dfrac{2}{5}-x\right|=0\Leftrightarrow\dfrac{2}{5}-x=0\Leftrightarrow x=\dfrac{2}{5}\)

Vậy \(Min_Q=\dfrac{9}{2}\) khi \(x=\dfrac{2}{5}\).

\(---\)

b) Ta thấy: \(\left|x+\dfrac{2}{3}\right|\ge0\forall x\)

\(\Rightarrow M=\left|x+\dfrac{2}{3}\right|-\dfrac{3}{5}\ge-\dfrac{3}{5}\forall x\)

Dấu \("="\) xảy ra khi: \(\left|x+\dfrac{2}{3}\right|=0\Leftrightarrow x+\dfrac{2}{3}=0\Leftrightarrow x=-\dfrac{2}{3}\)

Vậy \(Min_M=-\dfrac{3}{5}\) khi \(x=-\dfrac{2}{3}\).

\(---\)

c) Ta thấy: \(\left|\dfrac{7}{4}-x\right|\ge0\forall x\)

\(\Rightarrow-\left|\dfrac{7}{4}-x\right|\le0\forall x\)

\(\Rightarrow N=-\left|\dfrac{7}{4}-x\right|-8\le-8\forall x\)

Dấu \("="\) xảy ra khi: \(\left|\dfrac{7}{4}-x\right|=0\Leftrightarrow\dfrac{7}{4}-x=0\Leftrightarrow x=\dfrac{7}{4}\)

Vậy \(Max_N=-8\) khi \(x=\dfrac{7}{4}\).

a) Ta có: \(\left|\dfrac{2}{5}-x\right|\ge0\forall x\)

\(\Rightarrow Q=\dfrac{9}{2}+\left|\dfrac{2}{5}-x\right|\ge\dfrac{9}{2}\forall x\)

Dấu "=" xảy ra khi:

\(\dfrac{2}{5}-x=0\)

\(\Rightarrow x=\dfrac{2}{5}\)

Vậy: ... 

b) Ta có: \(\left|x+\dfrac{2}{3}\right|\ge0\forall x\)

\(\Rightarrow M=\left|x+\dfrac{2}{3}\right|-\dfrac{3}{5}\ge-\dfrac{3}{5}\)

Dấu "=" xảy ra:

\(x+\dfrac{2}{3}=0\)

\(\Rightarrow x=-\dfrac{2}{3}\)

Vậy: ...

c) Ta có: \(-\left|\dfrac{7}{4}-x\right|\le0\forall x\)

\(\Rightarrow N=-\left|\dfrac{7}{4}-x\right|-8\le-8\)

Dấu "=" xảy ra:

\(\dfrac{7}{4}-x=0\)

\(\Rightarrow x=\dfrac{7}{4}\)

Vậy: ...

`#\text{ID01}`

a)

`Q = 9/2 + |2/5 - x|`

Vì `|2/5 - x| \ge 0` `AA` `x`

`=> 9/2 + |2/5 - x| \ge 9/2` `AA` `x`

`=>` GTNN của Q là `9/2` khi `|2/5 - x| = 0`

`=> 2/5 - x = 0`

`=> x = 2/5`

b)

`M = |x + 2/3| - 3/5`

Vì `|x + 2/3| \ge 0` `AA` `x`

`=> |x + 2/3| - 3/5 \ge -3/5` `AA` `x`

`=>` GTNN của M là `-3/5` khi `|x + 2/3| = 0`

`=> x + 2/3 = 0`

`=> x = -2/3`

c)

`N=-|7/4 - x| - 8`

Vì `|7/4 - x| \ge 0` `AA` `x`

`=> -|7/4 - x| \le 0` `AA` `x`

`=> -|7/4 - x| - 8 \le -8` `AA` `x`

`=>` GTLN của N là `-8` khi `|7/4 - x| = 0`

`=> 7/4 - x = 0`

`=> x = 7/4`