Có \(\frac{18}{18+19+20}>\frac{18}{18+19+20+21}\)

\(\frac{19}{18+19+21}>\frac{19}{18+19+20+21}\)

\(\frac{20}{18+19+21}>\frac{20}{18+19+20+21}\)

\(\frac{21}{18+19+21}>\frac{21}{18+19+20+21}\)

=> \(\frac{18}{18+19+20}+\frac{19}{18+19+21}+\frac{20}{18+19+21}+\frac{21}{18+19+21}>\frac{18}{18+19+20+21}+\frac{19}{18+19+20+21}+\frac{20}{18+19+20+21}+\frac{21}{18+19+20+21}\)

=> \(\frac{18}{18+19+20}+\frac{19}{18+19+21}+\frac{20}{18+19+21}+\frac{21}{18+19+21}>\frac{18+19+20+21}{18+19+20+21}\)

=>\(\frac{18}{18+19+20}+\frac{19}{18+19+21}+\frac{20}{18+19+21}+\frac{21}{18+19+21}>1\)

=>M>1

Còn lại mình không biết, đúng thì tick nha

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

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

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

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

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

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

= \(\frac{20}{19}+\frac{20}{18}+...+\frac{20}{2}+\frac{20}{20}\)

=\(20.\left(\frac{1}{19}+\frac{1}{18}+...+\frac{1}{2}+\frac{1}{20}\right)\)

=\(20.\left(\frac{1}{20}+\frac{1}{19}+\frac{1}{18}+...+\frac{1}{2}\right)\)  

Vì tử số gấp 20 lần mẫu số nên phân số này bằng 20

\(B=\dfrac{20^{19}+1}{20^{20}+1}< \dfrac{20^{19}+1+19}{20^{20}+1+19}=\dfrac{20^{19}+20}{20^{20}+20}\)

\(B< \dfrac{20.\left(20^{18}+1\right)}{20.\left(20^{19}+1\right)}\)

\(B< \dfrac{20^{18}+1}{20^{19}+1}\)

\(B< A\)

\(\frac{20}{29}=\frac{40}{58}>\frac{29}{58}=\frac{1}{2}=\frac{20}{40}>\frac{19}{40}\)

\(\frac{19}{18}>1>\frac{20}{31}\)

=7¹⁸(7²+7-1)
=7¹⁸(49+7-1)
=7¹⁸ * 55
=7¹⁸ *5*11 chia hết cho 11
vậy ...

E = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 19x20

E x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 19x20x3

E x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 19x20x(21-18)

E x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 19x20x21 - 18x19x20.

E x 3 = 19x20x21

E = 19x20x21 : 3

E = 2660

\(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)

Biến đổi tử số 

\(19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}\)

= 1 + \(\left(1+\dfrac{18}{2}\right)+\left(1+\dfrac{17}{3}\right)+\left(1+\dfrac{16}{4}\right)+...+\left(1+\dfrac{1}{19}\right)\)

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

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

Vậy \(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)

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

Vậy A = 20

c.ơn nhìu a