Question

Tinh giá trị của biểu thức:\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{43.45}\)

Answer 1

\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{43.45}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{43.45}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{43}-\frac{1}{45}\)

\(=\frac{1}{3}-\frac{1}{45}=\frac{15}{45}-\frac{1}{45}=\frac{14}{45}\)

\(\Rightarrow A=\frac{14}{45}:2=\frac{14}{90}=\frac{7}{45}\)

Vậy \(A=\frac{7}{45}\).

Answer 2

Áp dụng công thức : \(\frac{1}{a}-\frac{1}{a+n}=\frac{n}{a\left(a+n\right)}\)

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{43\cdot45}\)

\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{43}-\frac{1}{45}\right)\)

\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{45}\right)\)

\(A=\frac{1}{2}\cdot\frac{14}{45}=\frac{7}{45}\)

Answer 3

A=\(\frac{7}{45}\)

Answer 4

A=7/15

Answer 5

Ta có A=1/3.5+1/5.7+1/7.9+...+1/43.45

        2A=2/3.5+2/5.7+2/7.9+...+2/43.45

        2A=1/3-1/5+1/5-1/7+1/7-1/9+...+1/43-1/45

        2A=1/3-1/45

        2A=15/45-1/45

        2A=14/45

  Vậy A=7/45

Answer 6

\(\frac{1}{2}A=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{43.45}\right)\)

\(\frac{1}{2}A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{43.45}\)

\(\frac{1}{2}A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{43}-\frac{1}{45}\)

\(\frac{1}{2}A=\frac{1}{3}-\frac{1}{45}=>A=...\)

Answer 7

\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{43.45}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{43}-\frac{1}{45}\)

\(2A=\frac{1}{3}-\frac{1}{45}=\frac{14}{45}\)

\(A=\frac{7}{45}\)