A = 1 12 + 1 13 + ... + 1 22 > 1 22 + 1 22 + ... + 1 22 ⏟ 11 s = 11 22 = 1 2

Làm luôn nhé

\(A=\frac{3}{8}+\frac{1}{5}+\frac{5}{6}>\frac{1}{6}+\frac{5}{6}=1\)

\(A=\frac{3}{8}+\frac{1}{5}+\frac{5}{6}< \frac{3}{8}+\frac{1}{4}+\frac{5}{4}=\frac{3}{8}+\frac{2}{8}+\frac{10}{8}=\frac{15}{8}< \frac{16}{8}=2\)

Vậy 1<A<2

\(B=\frac{5}{11}+\frac{5}{12}+\frac{5}{13}+\frac{5}{14}>\frac{5}{14}.4=\frac{10}{7}>1\)

\(B=\frac{5}{11}+\frac{5}{12}+\frac{5}{13}+\frac{5}{14}< \frac{5}{10}.4=2\)

Vậy 1<B<2

A = 1 12 + 1 13 + 1 14 + ... + 1 22 > 1 22 + 1 22 + ... 1 22 ⏟ 11 s = 11 22 = 1 2 .

A = 1 12 + 1 13 + 1 14 + ... + 1 22 > 1 2 ⇔ 1 12 + 1 13 + 1 14 + ... + 1 22 > 11 22 ⇔ 1 12 − 1 22 + 1 13 − 1 22 + 1 14 − 1 22 + ... + 1 22 − 1 22 > 0

Vì  1 12 > 0 , 1 13 > 0 , ... , 1 21 > 1 22 nên  1 12 − 1 22 > 0 , 1 13 − 1 22 > 0 , ... , 1 21 − 1 22 > 0 , 1 22 − 1 22 = 0

Suy ra  A > 1 2

a) A > 1 20 + 1 20 + ... + 1 20 ⏟ 10 s o = 10 20 = 1 2 .

b)  B = 1 5 + ... 1 9 + 1 10 + ... + 1 17 < 1 5 + ... + 1 5 ⏟ 5s o + 1 8 + ... + 1 8 ⏟ 8s o = 2

c)  C = 1 10 + 1 11 + 1 12 ... + 1 18 + 1 19 < 1 10 + 1 10 + ... 1 10 ⏟ 9 s o = 1

a)  A = 1 12 + 1 13 + 1 14 + ... + 1 22 > 1 22 + 1 22 + ... 1 22 ⏟ 11 s = 11 22 = 1 2 .

b) B = 1 6 + ... 1 9 + 1 10 + ... + 1 19 < 1 4 + ... + 1 4 ⏟ 4 s o + 1 10 + ... + 1 10 ⏟ 10 s o = 2

c)  C = 1 10 + 1 11 + ... + 1 100 > 1 10 + 1 100 = ... + 1 100 ⏟ 90 s o = 1 10 + 90 100 = 1

C = 1 10 + 1 11 + 1 12 + ... + 1 99 + 1 100 > 1 ⇔ C = 1 10 + 1 11 + 1 12 + ... + 1 20 + 1 21 + 1 22 + ... + 1 30 + ... + 1 91 + 1 92 + ... + 1 100 C > 1 10 + 10 20 + 10 30 + ... + 10 100 > 10 20 + 10 30 + 10 60 = 1 2 + 1 3 + 1 6 = 1

C = 1 10 + 1 11 + ... + 1 100 > 1 10 + 1 100 = ... + 1 100 ⏟ 90 s o = 1 10 + 90 100 = 1