Ta có : 

\(A=\frac{101}{1}+\frac{100}{2}+\frac{99}{3}+...+\frac{1}{101}\)

\(A=\left(101-1-...-1\right)+\left(\frac{100}{2}+1\right)+\left(\frac{99}{3}+1\right)+...+\left(\frac{1}{101}+1\right)\)

\(A=\frac{102}{102}+\frac{102}{2}+\frac{102}{3}+...+\frac{102}{101}\)

\(A=102\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}+\frac{1}{102}\right)\)

\(\Rightarrow\)\(\frac{A}{B}=\frac{102\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}}=\frac{102}{1}=102\)

Vậy \(\frac{A}{B}=102\)

Chúc bạn học tốt ~

\(VT=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{101}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{102}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{101}+\frac{1}{102}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{102}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}+\frac{1}{102}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{51}\)

\(=\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{102}\)

\(=VP\)

Bài 2:

M = 1/2.3/4.5/6...99/100

Ta có: \(\frac{a}{b}\) = 1 - \(\frac{b-a}{b}\) (a; b; n ∈ N* và b > a)

\(\frac{a+n}{b+n}\) = 1 - \(\frac{b-a}{b+n}\)

\(\frac{a}{b}\) < \(\frac{a+n}{b+n}\)

Áp dụng công thức trên ta có:

\(\frac12<\frac{1+1}{2+1}=\frac23\)

\(\frac34<\frac{3+1}{4+1}=\frac45\)

\(\frac56\) < \(\frac{5+1}{6+1}\) = \(\frac67\)

............................

\(\frac{99}{100}\) < \(\frac{99+1}{100+1}\) = \(\frac{100}{101}\)

Cộng vế với vế ta có:

M = \(\frac12\).\(\frac34\).\(\frac56\)...\(\frac{99}{100}\) < \(\frac23\).\(\frac45\)..\(\frac{100}{101}\) = N

M < N (đpcm)

b; M.N = \(\frac12\).\(\frac34\).\(\frac56\)...\(\frac{99}{100}\).\(\frac23\).\(\frac45\)..\(\frac{100}{101}\)

M.N = \(\frac{1.3.5\ldots99}{3.5\ldots101}\). \(\frac{2.4.6\ldots100}{2.4.6\ldots100}\)

M.N = 1/100.101

a) = 3/3 x ( -24/54 +45/54 ) : 7/12

   = 1 x 21/54 x 12/7

   = 18/27 

( hiện tại mik chỉ lm đc thế này thui. thông cảm nk )

S = \(\frac12\times\frac13\) + \(\frac13\times\frac14\) + \(\frac14\times\frac15\) + \(\frac15\times\frac16\) + \(\frac17\times\frac18\) + \(\frac18\times\frac19\)

S = \(\frac12\) - \(\frac13\) + \(\frac13\) - \(\frac14\) + \(\frac14\) - \(\frac15\) + \(\frac15\) - \(\frac16\) + \(\frac17\) - \(\frac18\) + \(\frac18\) - \(\frac19\)

S = \(\frac12\) - \(\frac19\)

S = \(\frac{9}{18}-\frac{2}{18}\)

S = \(\frac{7}{18}\)

a) \(\frac{53}{101}\cdot-\frac{13}{97}+\frac{53}{101}\cdot-\frac{84}{97}\)

\(=\frac{53}{101}\cdot\left(-\frac{13}{97}-\frac{84}{97}\right)\)

\(=\frac{53}{101}\cdot\left(-1\right)\)

\(=-\frac{53}{101}\)

giống huy nhé

Ta có:

\(A=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}>\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}=\frac{100}{200}=\frac{1}{2}\)

\(\Rightarrow A=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}>\frac{1}{2}\)