Question

[superman]

Tìm x biết : \(\left(\frac{1}{1\cdot2}+\frac{1}{1\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99.100}\right)-2\cdot x=\frac{1}{2}\).

Các bạn giúp mình với .

Answer 1

[Tẫn]

\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+........+\frac{1}{99\cdot100}\right)-2x=\frac{1}{2}\)

\(\left(\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+...+\frac{100-99}{99\cdot100}\right)-2x=\frac{1}{2}\)

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

\(\left(1-\frac{1}{100}\right)-2x=\frac{1}{2}\)

\(\frac{99}{100}-2x=\frac{1}{2}\)

\(2x=\frac{99}{100}-\frac{1}{2}\)

\(2x=\frac{49}{100}\)

\(x=\frac{49}{100}:2\)

\(x=\frac{49}{200}\)

Answer 2

[Phạm Tuấn Đạt]

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

\(\frac{99}{100}-2x=\frac{1}{2}\)

\(\frac{99-50}{100}=2x\)

\(49=200x\)

\(x=\frac{49}{200}\)