Question

Tính:

A= \(\frac{1}{2.5}\)+\(\frac{1}{5.8}\)+\(\frac{1}{8.11}\)+....+\(\frac{1}{92.95}\)+\(\frac{1}{95.98}\).

Answer 1

A=1/3x(1/2x5+1/5x8+......+1/95x98)

A=1/3x(1/2-1/5+1/5-1/8+.........+1/95-1/98)

A=1/3x(1/2-1/98)

A=1/3x24/49

A=8/49

Answer 2

A =\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)

A = \(\frac{1.3}{2.5.3}+\frac{1.3}{5.8.3}+\frac{1.3}{8.11.3}+...+\frac{1.3}{92.95.3}+\frac{1.3}{95.98.3}\)

A = \(\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)

A =\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)

A =\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)

A =\(\frac{1}{3}.\frac{97}{98}\)

A =\(\frac{97}{294}\)

Answer 3

bạn Nguyễn Hoàng Phúc sai ở chỗ 1/2-1/98

Answer 4

Dễ chứng minh:\(\frac{n}{k.\left(k+n\right)}=\frac{1}{k}-\frac{1}{k+n}\)

áp dụng ta có:

3A=\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.5}+...+\frac{3}{92.95}+\frac{3}{95.98}\)

    =\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\)

    =\(\frac{1}{2}-\frac{1}{98}\)

    =\(\frac{24}{49}\)

3A=\(\frac{24}{49}\)

A=\(\frac{8}{49}\)

Answer 5

Mình nhầm sửa lại chỗ dấu bằng thứ 6

\(\frac{1}{3}.\frac{24}{49}=\frac{8}{49}\)