Ta có tính chất `|P|>=P,|P|>=-P`

`=>{(|x-2|>=x-2),(|x-4|>=4-x):}`

`=>B>=x-2+4-x=2`

Dấu "=" xảy ra khi `{(x-2>=0),(x-4<=0):}`

`<=>{(x>=2),(x<=4):}`

`<=>2<=x<=4`

\(2\left|x+1\right|+\left|2x-3\right|\)

\(=\left|2x+2\right|+\left|2x-3\right|\)

\(=\left|2x+2-2x+3\right|\ge5\)

\(A_{min}=5\)

Amin= 5 ? khi đó x bằng mấy?

\(a,A=\left|2-4x\right|-6\ge-6\\ A_{min}=-6\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\\ b,x^2+1\ge1\Leftrightarrow B=1-\dfrac{4}{x^2+1}\ge1-\dfrac{4}{1}=-3\\ B_{min}=-3\Leftrightarrow x=0\)

A= |2x+5| + |2x-1|

=> A=|2x+5| + |1-2x|

Ap dụng tính chất:    |A| \(\ge\)A. Dấu = xảy ra khi A\(\ge\)0

=> |2x+5| \(\ge\)2x+5. Dấu = xảy ra khi 2x+5\(\ge\)0       (1)

     |1-2x| \(\ge\)1-2x. Dấu = xảy ra khi 1-2x\(\ge\)0          (2)

=> A\(\ge\)2x+5+1-2x. Dấu = xảy ra khi dấu = ở (1);(2) đồng thời xảy ra

=>\(\ge\)6

=>  GTNN của A là 6 <=> x=0

   Vậy Min A=6 <=> x=0

Biểu thức này không có min và cũng không có max

a) Ta có: \(A=x^2-5x+7\)

\(=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)

b) Ta có: \(B=2x^2-8x+15\)

\(=2\left(x^2-4x+\dfrac{15}{2}\right)\)

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

\(=2\left(x-2\right)^2+7\ge7\forall x\)

Dấu '=' xảy ra khi x=2

a. `A=x^2-5x+7`

`=x^2-2.x. 5/2 + (5/2)^2 +3/4`

`=(x-5/2)^2 + 3/4`

`=> A_(min) =3/4 <=> x-5/2 =0<=>x=5/2`

b) `B=2x^2-8x+15`

`=[(\sqrt2x)^2 -2.\sqrt2 x . 2\sqrt2 +(2\sqrt2)^2] +7`

`=(\sqrt2x-2\sqrt2)^2+7`

`=> B_(min)=7 <=> x=2`.

a) \(A=x^2-5x+7\)

\(=x^2-2.\dfrac{5}{2}x+\left(\dfrac{5}{2}\right)^2+\dfrac{3}{4}\)

\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\)

Mặt khác, ta có \(\left(x-\dfrac{5}{2}\right)^2\ge0\forall x\)  \(\Rightarrow\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

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

Vậy \(A_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{5}{2}\)

b) \(B=2x^2-8x+15\)

\(=4x^2-2.2x.2+2^2+11\)

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

Vì \(\left(2x-2\right)^2\ge0\forall x\) nên \(\left(2x-2\right)^2+11\ge11\forall x\)

Dấu "=" xảy ra khi  \(\left(2x-2\right)^2=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)

Vậy \(B_{min}=11\) khi \(x=1\)

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

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

b) \(B=3x^2+4x-13=3\left(x^2+\dfrac{4}{3}x+\dfrac{4}{9}\right)-\dfrac{35}{3}=3\left(x+\dfrac{2}{3}\right)^2-\dfrac{35}{3}\ge-\dfrac{35}{3}\)

\(minB=-\dfrac{35}{3}\Leftrightarrow x=-\dfrac{2}{3}\)

a: Ta có: \(N=-x^2-x-1\)

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

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

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

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

b: ta có: \(B=3x^2+4x-13\)

\(=3\left(x^2+\dfrac{4}{3}x-\dfrac{13}{3}\right)\)

\(=3\left(x^2+2\cdot x\cdot\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{43}{9}\right)\)

\(=3\left(x+\dfrac{2}{3}\right)^2-\dfrac{43}{3}\ge-\dfrac{43}{3}\forall x\)

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