Giải ra giúp tôi với mấy thánh

Xét tử:

tử = 1/18 + 2/17 + 3/16 + ... + 18/1 + (1+1+1+...+1)(18 số 1)

    =(1/18 + 1)+(2/17 + 1)+...+(18/1 + 1)

    =19/18 + 19/17 + ... + 19/1

    =19(1/18 + 1/17 + ... + 1/1)

Nên tử/ mẫu =19

Xét tử số 

1/18+2/17+3/16+...+18/1+18

=[(1/18)+1]+[(2/17)+1]+[(3/16)+1]+...+[(18/1)+1]

=19/18+19/17+19/16+...+19/1

=19.[(1/18)+(1/17)+(1/16)+...+1/1]

=>phân số trên bằng 19

\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{\dfrac{19}{1}+\dfrac{18}{2}+\dfrac{17}{3}+....+\dfrac{1}{19}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{1+\left(\dfrac{18}{2}+1\right)+\left(\dfrac{17}{3}+1\right)+\left(\dfrac{1}{19}+1\right)}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{1+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{20}{19}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{20.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}\)

\(=\dfrac{1}{20}\)

a ) 11/125

b ) 1

\(A=\dfrac{1}{20}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{19\cdot20}\right)\)

\(=\dfrac{1}{20}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)

=1/20-19/20=-18/20=-9/10

a) \(\dfrac{17}{20}< \dfrac{18}{20}< \dfrac{18}{19}\Rightarrow\dfrac{17}{20}< \dfrac{18}{19}\)

b) \(\dfrac{19}{18}>\dfrac{19+2024}{18+2024}=\dfrac{2023}{2022}\Rightarrow\dfrac{19}{18}>\dfrac{2023}{2022}\)

c) \(\dfrac{135}{175}=\dfrac{27}{35}\)

\(\dfrac{13}{17}=\dfrac{26}{34}< \dfrac{26+1}{34+1}=\dfrac{27}{35}\)

\(\Rightarrow\dfrac{13}{17}< \dfrac{135}{175}\)

\(\left(\dfrac{5}{2}\right)^{18}:\left(1+\dfrac{3}{2}\right)^{17}\)

\(=\left(\dfrac{5}{2}\right)^{18}:\left(\dfrac{5}{2}\right)^{17}\)

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

\(\left(\dfrac{5}{2}\right)^{18}:\left(1+\dfrac{3}{2}\right)^{17}\)

\(=\left(\dfrac{5}{2}\right)^{18}:\left(\dfrac{2}{2}+\dfrac{1}{2}\right)^{17}\)

\(=\left(\dfrac{5}{2}\right)^{18}:\left(\dfrac{5}{2}\right)^{17}\)

\(=\left(\dfrac{5}{2}\right)^{18-17}\)

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