Question

A=\(\frac{10}{1.2}+\frac{10}{2.3}+\frac{10}{3.4}+...+\frac{10}{98.99}+\frac{10}{99.100}\)

Answer 1

\(A=\frac{10}{1.2}+\frac{10}{2.3}+\frac{10}{3.4}+...+\frac{10}{98.99}+\frac{10}{99.100}\)

\(A=10.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(A=10.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=10.\left(\frac{1}{1}-\frac{1}{100}\right)\)

\(A=10.\frac{99}{100}\)

\(A=\frac{99}{10}\)

Học tốt 

Answer 2

a= \(10.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

a=\(10.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

a=\(10.\left(1-\frac{1}{100}\right)\)

a=\(\frac{99}{10}\)

Answer 3

\(A=\frac{10}{1\cdot2}+\frac{10}{2\cdot3}+\frac{10}{3\cdot4}+...+\frac{10}{99\cdot100}\)

\(A=10\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\right)\)

\(A=10\left(1-\frac{1}{100}\right)\)

\(A=10\cdot\frac{99}{100}\)

\(A=\frac{99}{10}\)

Answer 4

len rank dai ky tuong chua may thang nhoc