Question

1.Tính giá trị biểu thức sau:

A=\(\dfrac{2}{2.5}\)\(+\)\(\dfrac{2}{5.8}\)\(+\)\(\dfrac{2}{8.11}\)\(+\)...........\(+\)\(\dfrac{2}{92.95}\)\(+\)\(\dfrac{2}{95.98}\)

Giúp mình nha các cậu

Answer 1

\(A=\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{92.95}+\dfrac{2}{95.98}\)

\(A=2\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}+\dfrac{1}{95.98}\right)\)

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

\(A=\dfrac{2}{2}\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\)

\(A=\dfrac{24}{49}\)

Answer 2

\(A=2.\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{95.98}\right)\)

\(A=\dfrac{2}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+....+\dfrac{3}{95.98}\right)\)

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

\(A=\dfrac{2}{3}\dfrac{24}{49}=\dfrac{16}{49}\)

Answer 3

Ta có: A=\(\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}\)

\(\Rightarrow A=\dfrac{3}{2}.\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{95.98}\right)\)\(\Rightarrow A=\dfrac{3}{2}.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}\right)\)\(\Rightarrow A=\dfrac{3}{2}.\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\)

\(\Rightarrow A=\dfrac{3}{2}.\left(\dfrac{49}{98}-\dfrac{1}{98}\right)\)

\(\Rightarrow A=\dfrac{3}{2}.\dfrac{48}{98}\)

\(\Rightarrow A=\dfrac{3.2.2.12}{2.2.49}\)

\(\Rightarrow A=\dfrac{36}{49}\)

Answer 4

A= \(\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+.....+\dfrac{2}{92.95}+\dfrac{2}{95.98}\)

= \(2.\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+....+\dfrac{1}{92.95}+\dfrac{1}{95.98}\right)\)

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

= \(2.\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\)

= \(2.\dfrac{24}{49}\)

= \(\dfrac{48}{49}\)

Answer 5

Mk làm bị thiếu cái này ms đúng.

A= \(\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{92.95}+\dfrac{2}{95.98}\)

= \(2.\dfrac{1}{3}.\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}+\dfrac{1}{95.98}\right)\)

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

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

= \(\dfrac{2}{3}.\dfrac{24}{49}\)

= \(\dfrac{16}{49}\)