a: Ta có: \(A=x^2+3x+4\)

\(=x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{7}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

a) \(2x^2-x+1=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{4}\)

b) \(5x-x^2+4=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{41}{4}\le\dfrac{41}{4}\)

\(ĐTXR\Leftrightarrow x=\dfrac{5}{2}\)

c) \(x^2+5y^2-2xy+4y+3=\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\)

\(ĐTXR\Leftrightarrow\)\(x=y=-\dfrac{1}{2}\)

b: ta có: \(-x^2+5x+4\)

\(=-\left(x^2-5x-4\right)\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{41}{4}\right)\)

\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{41}{4}\le\dfrac{41}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)

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`

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

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 \(A_{max}=4;A_{min}=-1\)

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

Câu 2:

ĐKXĐ: x<>0

\(B=\dfrac{-x^2-x-1}{x^2}\)

\(=-1-\dfrac{1}{x}-\dfrac{1}{x^2}\)

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

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

\(=-\left(\dfrac{1}{x}+\dfrac{1}{2}\right)^2-\dfrac{3}{4}< =-\dfrac{3}{4}\forall x< >0\)

Dấu '=' xảy ra khi 1/x+1/2=0

=>1/x=-1/2

=>x=-2

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

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

B=y^2-y+1

=y^2-2*y*1/2+1/4+3/4

=(y-1/2)^2+3/4>=3/4

Dấu = xảy ra khi y=1/2

E=-x^2+x+2

=-(x^2-x-2)

=-(x^2-x+1/4-9/4)

=-(x-1/2)^2+9/4<=9/4

Dấu = xảy ra khi x=1/2

 ĐKXĐ :\(x\ge0\)

\(x-4\sqrt{x}+5\)

\(=x-4\sqrt{x}+4+1\)

\(=\left(\sqrt{x}-2\right)^2+1\ge1\forall x\ge0\)

Dấu"=" xả ra <=> \(\left(\sqrt{x}-2\right)^2=0\)

                    \(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)