Só rì ae, ko có CMR A

a) Ta có:

(n-1)/n < n/(n+1)

vì (n-1).(n+1)=n²-1 < n²

=>

1/2 < 2/3

3/4 < 4/5

....

99/100 < 100/101

Vậy A < B

b). Ta lại có:

A.B = 1/2 . 2/3 . 3/4 . 4/5 .... . 99/100 . 100/101 = 1/100

Mà A<B => A.A<A.B=1/100

=> A² < 1/100

=> A < 1/10<1

A < 1/10<1

A < 1/10<1

\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{6}-\dfrac{1}{7}=\dfrac{1}{2}-\dfrac{1}{7}=\dfrac{5}{14}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}=\dfrac{21-2}{42}=\dfrac{19}{42}\)

Lời giải:
Gọi biểu thức số 1 là A và số 2 là B

\(A=\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+\frac{5-4}{4\times 5}+\frac{6-5}{5\times 6}+\frac{7-6}{6\times 7}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)

B tương tự A:
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{20}-\frac{1}{21}\)

\(=\frac{1}{2}-\frac{1}{21}=\frac{19}{42}\)

\(A=\frac{1}{2\times3}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{98\times99}\)

\(=\frac{1}{6}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}+...+\frac{1}{98\times99}\)

\(=\frac{1}{6}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{99}\)

\(=\frac{1}{6}+\frac{1}{4}-\frac{1}{99}=\frac{161}{396}>\frac{160}{400}=\frac{2}{5}\)