A=1/1.2+1/12+...+1/99.100

A=7/12+...1/99.100

Suy ra A>7/12 (1)

A=1-1/2+1/3-1/4+...+1/99-1/100

A=(1/2+1/3)-(1/4-...+1/100)

A=5/6-(1/4-...+1/100)

suy ra A<5/6 (2)

Vậy 7/12<A<5/6

chắc chắn đúng

Lê Tùng lâm bài của bạn chưa đúng vì

A = \(\frac{1}{1.2}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)

Chứ không phải là: \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{98.99}+\frac{1}{99.100}\)

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)

1/51+1/52+1/53 +...+1/100

Dễ mà bạn.

Ta có :

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

cảm ơn bạn nha

huj nãy sai đề hẻn chj ko làm dc

Ta có : \(\frac{\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}}{\frac{6}{5}+\frac{6}{7}-\frac{2}{3}+\frac{6}{11}}=\frac{\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}}{2\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}\right)}=\frac{1}{2}\)

Lại có : \(\frac{\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right).2021}{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}}=\frac{0.2021}{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}}=0\)

Khi đó \(B=\frac{1}{2}+0=\frac{1}{2}\)

\(\Rightarrow A=5\left(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99x100}\right)\)

\(\Rightarrow A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Rightarrow A=5\left(1-\frac{1}{100}\right)\)

\(\Rightarrow A=\frac{5x99}{100}=\frac{99}{20}\)

\(A=\frac{5}{1}-\frac{5}{2}+\frac{5}{2}-\frac{5}{3}+\frac{5}{3}-\frac{5}{4}+....+\frac{5}{99}-\frac{5}{100}\)

\(A=\frac{5}{1}+\left(-\frac{5}{2}+\frac{5}{2}\right)+\left(-\frac{5}{3}+\frac{5}{3}\right)+\left(-\frac{5}{4}+\frac{5}{4}\right)+...\left(-\frac{5}{99}+\frac{5}{99}\right)+\frac{5}{100}\)

\(A=\frac{5}{1}+0+0+....+0+\frac{5}{100}\)

\(A=\frac{500}{100}+\frac{5}{100}=\frac{205}{100}=\frac{101}{20}\)

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