Question

[tran viet duc]

1/4.8+1/8.12+1/12.16+......+1/2008.2012?

Answer 1

[迪丽热巴·迪力木拉提]

Gọi biểu thức cần tình giá trị là A.

Ta có: \(A=\dfrac{1}{4.8}+\dfrac{1}{8.12}+\dfrac{1}{12.16}+...+\dfrac{1}{2008.2012}\)

\(\Leftrightarrow4A=4\left(\dfrac{1}{4.8}+\dfrac{1}{8.12}+\dfrac{1}{12.16}+...+\dfrac{1}{2008.2012}\right)\)

\(\Leftrightarrow4A=\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{16}+...+\dfrac{1}{2008}-\dfrac{1}{2012}\)

\(\Leftrightarrow4A=\dfrac{1}{4}-\dfrac{1}{2012}=\dfrac{251}{1006}\)

\(\Leftrightarrow A=\dfrac{\dfrac{251}{1006}}{4}=\dfrac{251}{4024}\)

Answer 2

[|THICK TUNA|]

Ta gọi Biểu thức trên là A ta có:

A= \(\dfrac{1}{4.8}+\dfrac{1}{8.12}+\dfrac{1}{12.16}+...+\dfrac{1}{2008.2012}\)

4.A=\(\dfrac{4}{4.8}+\dfrac{4}{8.12}+\dfrac{4}{12.16}+...+\dfrac{4}{2008.201}\)

4.A=\(\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{16}+...+\dfrac{1}{2008}-\dfrac{1}{2012}\)

4.A=\(\dfrac{1}{4}-\dfrac{1}{2012}\)

4.A=\(\dfrac{502}{2012}\)

  A=\(\dfrac{502}{2012}:4\)

  A=\(\dfrac{502}{8048}\)

  A=\(\dfrac{251}{4024}\)

  Vậy biểu thức trên có giá trị là \(\dfrac{251}{4024}\)