Question

[Phạm Tường Nhật]

Tìm x biết: \(\left(\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\right).x=1\)

Answer 1

[Nga Nguyễn]

Đặt \(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}.\)

\(2A=\frac{2}{2.3}+\frac{2}{6.5}+\frac{2}{10.7}+...+\frac{2}{198.101}\)

\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(4A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{1}-\frac{1}{101}\)\(=\frac{100}{101}\)

\(A=\frac{100}{101}\div4=\frac{25}{101}\)

\(\Rightarrow\frac{25}{101}x=1\)

\(x=1\div\frac{25}{101}=\frac{101}{25}\)

Answer 2

[nguyenthutrang]

101phan 25