a, Với 3 =< x =< 5 

\(C=x-3+5-x=2\)

Với x >= 5 

\(C=x-3+x-5=2x-8\)

Với x =< 3 

\(C=3-x+5-x=-2x+8\)

b, Với x >= 7/3 

\(D=3x-7+3x+5=6x-2\)

Với x =< -5/3 

\(D=7-3x-3x-5=-6x+2\)

Với -5/3 =< x =< 7/3 

\(D=7-3x+3x+5=12\)

THAM KHẢO Ạ! 

a) C = |x-3| + |5-x|

Khi x <= 3:

- |x-3| = 3-x (với x <= 3)

- |5-x| = 5-x

Do đó, C = (3-x) + (5-x) = 8 - 2x

Khi 3 < x <= 5:

- |x-3| = x-3 (với 3 < x <= 5)

- |5-x| = 5-x

Do đó, C = (x-3) + (5-x) = 2

Khi x > 5:

- |x-3| = x-3

- |5-x| = x-5

Do đó, C = (x-3) + (x-5) = 2x - 8

b) D = |3x-7| + |3x+5|

Khi 3x-7 >= 0 (tức x >= 7/3) và 3x+5 >= 0 (tức x >= -5/3):

- D = (3x-7) + (3x+5) = 6x - 2

Khi 3x-7 < 0 và 3x+5 >= 0 (tức -5/3 < x <= 7/3):

- D = -(3x-7) + (3x+5) = 12

Khi 3x-7 < 0 (tức x < 7/3) và 3x+5 < 0 (tức x < -5/3):

- D = -(3x-7) + -(3x+5) = -6x - 12

tìm gtnn đk bn 

a, \(\left|x-2\right|\ge0\Rightarrow A=\left|x-2\right|+7\ge7\)

Dấu ''='' xảy ra khi x = 2 

b, \(\left|x+3\right|\ge0\Rightarrow3\left|x+3\right|+9\ge9\)

Dấu ''='' xảy ra khi x = - 3 

59 + 56 + 21 + 14 + 12

= (59 + 21) + (56 + 14) + 12

= 80 + 70 + 12

= 150 + 12

= 162

\(59\) \(+\) \(56\) \(+\) \(21\) \(+\) \(14\) \(+\) \(12\)

\(=\) \(\left(59+21\right)\) \(+\) \(\left(56+14\right)\) \( +\) \(12\)

\(=\) \(80\) \(+\) \(70\) \(+\) \(12\)

\(=\) \(150\) \(+\) \(12\)

\(=\) \(162\)

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2023\cdot2024}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\\ =1-\dfrac{1}{2024}\\ =\dfrac{2023}{2024}\\ B=\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+...+\dfrac{4}{2022\cdot2024}\\ =2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2022\cdot2024}\right)\\ =2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2022}-\dfrac{1}{2024}\right)\\ =2\left(\dfrac{1}{2}-\dfrac{1}{2024}\right)\\ =2\left(\dfrac{1012-1}{2024}\right)=\dfrac{1011}{1012}\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2023.2024}\\ A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2023}-\dfrac{1}{1024}\\ A=1-\dfrac{1}{2024}=\dfrac{2023}{2024}\)

\(B=\dfrac{4}{2.4}+\dfrac{4}{4.6}+...+\dfrac{4}{2022.2024}\\ B=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2022}-\dfrac{1}{2024}\right)\\ B=2\left(\dfrac{1}{2}-\dfrac{1}{2024}\right)=\dfrac{2.1011}{2024}=\dfrac{1011}{1012}\)

sửa đề ý B nhé

a) 

\(3,5x-2=\dfrac{1}{3}x+\dfrac{2}{7}\\ \Leftrightarrow\dfrac{7}{2}x-\dfrac{1}{3}x=\dfrac{2}{7}+2\\ \Leftrightarrow x\left(\dfrac{7}{2}-\dfrac{1}{3}\right)=\dfrac{16}{7}\\ \Leftrightarrow\dfrac{19}{6}\cdot x=\dfrac{16}{7}\\ \Leftrightarrow x=\dfrac{16}{7}:\dfrac{19}{6}\\ \Leftrightarrow x=\dfrac{96}{133}\)

b) 

\(2\left[\dfrac{x-1}{40}-3\left(x-1\right)\right]=2\\ \Leftrightarrow\dfrac{x-1}{40}-3\left(x-1\right)=1\\ \Leftrightarrow\dfrac{x-1-120\left(x-1\right)}{40}=1\\ \Leftrightarrow x-1-120x+120=40\\ \Leftrightarrow-119x+119=40\\ \Leftrightarrow-119x=40-119=-79\\ \Leftrightarrow x=\dfrac{79}{119}\)

a, \(3,5x-2=\dfrac{1}{3}x+\dfrac{2}{7}\Leftrightarrow\dfrac{19}{6}x=\dfrac{2}{7}+2=\dfrac{16}{7}\Leftrightarrow x=\dfrac{16}{7}:\dfrac{19}{6}=\dfrac{96}{133}\)

b, \(2\left[x-\dfrac{1}{40}-3\left(x-1\right)\right]=2\)

\(\Leftrightarrow x-\dfrac{1}{40}-3x+3=1\Leftrightarrow-2x=1+\dfrac{1}{40}-3=-\dfrac{79}{40}\Leftrightarrow x=\dfrac{79}{80}\)

\(\left(\dfrac{1}{3}\right)^2\cdot\dfrac{1}{3}\cdot9^2\\ =\left(\dfrac{1}{3}\right)^2\cdot\dfrac{1}{3}\cdot\left(3^2\right)^2\\ =\left(\dfrac{1}{3}\right)^{1+2}\cdot3^{2\cdot2}\\ =\left(\dfrac{1}{3}\right)^3\cdot3^4\\ =\dfrac{1}{3^3}\cdot3^4\\ =\dfrac{3^4}{3^3}\\ =3\)

\(\left(\dfrac{1}{3}\right)^2.\dfrac{1}{3}.9^2=\dfrac{1}{9}.\dfrac{1}{3}.81=\dfrac{81}{27}=\dfrac{9}{3}=3\)

Ta có: 

\(-\dfrac{7}{9}=\dfrac{-14}{18}\)

\(-\dfrac{5}{6}=\dfrac{-15}{18}\)

\(\dfrac{-2}{9}=\dfrac{-4}{18}\)

\(\dfrac{4}{3}=\dfrac{24}{18}\)

\(\Rightarrow\dfrac{-15}{18}< \dfrac{-14}{18}< \dfrac{-4}{18}< 0< \dfrac{24}{18}\)

\(\Rightarrow-\dfrac{5}{6}< -\dfrac{7}{9}< -\dfrac{2}{9}< 0< \dfrac{4}{3}\)

Sắp xếp: \(\dfrac{4}{3};0;-\dfrac{2}{9};-\dfrac{7}{9};-\dfrac{5}{6}\)

\(\dfrac{2^2\cdot4\cdot32}{2^2\cdot2^5}\\ =\dfrac{2^2\cdot2^2\cdot2^5}{2^2\cdot2^5}\\ =\dfrac{2^{2+2+5}}{2^{2+5}}\\ =\dfrac{2^9}{2^7}\\ =2^{9-7}\\ =2^2\)