A=0 nhé

a) \(\left|x-7\right|+\left|x+5\right|=\left|7-x\right|+\left|x+5\right|\ge\left|7-x+x+5\right|=12\)

Dấu "=" xảy ra khi \(-5\le x\le7\)

b) Đặt \(\left|2x-1\right|=t\left(t\ge0\right)\)

ta được \(t^2-3t+2=\left(t^2-2.\frac{3}{2}.x+\frac{9}{4}\right)-\frac{1}{4}=\left(t-\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Dấu "=" xảy ra khi \(\left(t-\frac{3}{2}\right)^2=0\Leftrightarrow t-\frac{3}{2}=0\Leftrightarrow t=\frac{3}{2}\Leftrightarrow\left|2x-1\right|=\frac{3}{2}\)

<=>\(\orbr{\begin{cases}2x-1=-\frac{3}{2}\\2x-1=\frac{3}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-\frac{1}{2}\\2x=\frac{5}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=\frac{5}{4}\end{cases}}\)

Vậy...........

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

\(\Rightarrow A=\left|2x-1\right|+\left|3-2x\right|\ge\left|2x-1+3-2x\right|\)

\(\Rightarrow A=\left|2x-1\right|+\left|3-2x\right|\ge\left|2\right|=2\)

dấu "="xảy ra khi \(\left(2x-1\right).\left(3-2x\right)\ge0\)

\(\Rightarrow\frac{1}{2}\le x\le\frac{3}{2}\)

vậy min A=2 khi \(\frac{1}{2}\le x\le\frac{3}{2}\)

Giá trị tuyệt đối à

Đặt A = |2x + 5| + |2x - 7|

=>A = |2x + 5| + |7 - 2x| \(\ge\)|2x + 5 + 7 - 2x| = |12| = 12

Dấu "=" xảy ra <=> (2x + 5)(7 - 2x) \(\ge\)0

=> -5/2 \(\le\)x \(\le\)7/2

Vậy MinA = 12 <=> -5/2 \(\le\)x \(\le\)7/2

Ta co: |1-2x|>hoac bang 0

suy ra 3. |1-2x|>hoac bang 0

3. |1-2x|-5> hoac bang -5

Dau "='' xay ra khi 1-2x=0 hay x=1/2

Vay Min A=-5 khi x= 1/2

Áp dụng BĐT /A/+/B/\(\ge\)/A+B/

\(\Leftrightarrow\)/2x-3/+/7-2x/\(\ge\)/2x-3+7-2x/=4.min

Vậy min là 4.dấu = xảy ra khi \(x=\frac{3}{2}\)

Ta có: |2x - 3| lớn hơn hoặc bằng 2x - 3

          |7 - 2x| lớn hơn hoặc bằng 7 - 2x

=> P lớn hơn hoặc bằng (2x - 3) + (7 - 2x) = 4