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![](https://rs.olm.vn/images/avt/0.png?1311)
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= 1/2-1/3+1/3-1/4+...+ 1/19-1/20
= 1/2-1/20
=9/20
có phải như thế này ko bn
\(A=\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{19.20}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{19}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}\)
A = \(\frac{9}{20}\)
\(B=\frac{1}{99.100}-\frac{1}{98.99}-\frac{1}{97.98}-.....-\frac{1}{1.2}=-\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\right)\)
\(B=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=-\left(1-\frac{1}{100}\right)\)
B = \(-\frac{99}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)=\frac{1}{100}+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow x-\frac{98}{99}=\frac{1}{99}\Leftrightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=\)\(1-\frac{1}{100}\)
\(=\)\(\frac{99}{100}\)
Vậy \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}=\frac{99}{100}\)
Chúc bạn học tốt ~
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
ĐÚNG 100%
![](https://rs.olm.vn/images/avt/0.png?1311)
Làm tiếp
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...........+\frac{1}{99}-\frac{1}{100}\)
A=\(1-\frac{1}{100}\)
A=\(\frac{100}{100}-\frac{1}{100}\)
A=\(\frac{99}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+..........+\frac{1}{49.50}\)
\(\Leftrightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{49}-\frac{1}{50}\)
\(\Leftrightarrow A=1-\frac{1}{50}=\frac{49}{50}\)
cái kia tự tìm
C=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+....+\(\frac{1}{97.98}\)+\(\frac{1}{98.99}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu hỏi của lương hiếu - Toán lớp 6 - Học toán với OnlineMath
Làm như link trên nhưng bỏ hạng tử \(\frac{1}{99.100}\)đi
Bước cuối: \(1-\frac{1}{99}=\frac{98}{99}\)
\(a,\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(b,\frac{x}{y}=\frac{3}{5}\)
\(\Leftrightarrow\frac{x}{3}=\frac{y}{5}\)
\(\text{Áp dụng tính chất dãy tỉ số bằng nhau ta có :}\)
\(\frac{x}{3}=\frac{y}{5}=\frac{x+y}{3+5}=\frac{18}{8}=\frac{9}{4}\)
\(\Rightarrow\frac{x}{3}=\frac{9}{4}\Rightarrow x=\frac{27}{4}\)
\(\frac{y}{5}=\frac{9}{4}\Rightarrow y=\frac{45}{4}\)