Gọi biểu thức cần tình giá trị là A.

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

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

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

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

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

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 \(\dfrac{251}{4024}\)

1/6 uyhvnb

Ý ban là:

\(\frac{1}{6}\)

a: \(=\left(\dfrac{28}{42}+\dfrac{12}{42}-\dfrac{3}{42}\right):\left(\dfrac{-28}{28}-\dfrac{12}{28}+\dfrac{3}{28}\right)\)

\(=\dfrac{37}{42}:\dfrac{-37}{28}=\dfrac{-28}{42}=-\dfrac{2}{3}\)

b: \(=\dfrac{2+8+18+32+50}{12+48+108+192+300}=\dfrac{110}{660}=\dfrac{1}{6}\)

= \(\frac{1.2+2.4+3.6+4.8+5.10}{3.2\left(1.2+2.4+3.6+4.8+5.10\right)}=\frac{1}{6}\)
nha :*

cách giải nè.

Ta có: \(\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)

=\(\frac{1.2+2.4+3.6+4.8+5.10}{3.2.1.2+3.2.2.4+3.2.3.6+3.2.4.8+3.2.5.10}\)

=\(\frac{1.2+2.4+3.6+4.8+5.10}{3.2.\left(1.2+2.4+3.6+4.8+5.10\right)}\)

=\(\frac{1}{6}\)

\(\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)

\(=\frac{1.2+2.2.2+3.3.2+4.4.2+5.5.2}{3.4+2.3.2.4+3.3.3.4+4.3.4.4+5.3.5.4}\)

\(=\frac{2.\left(1+2.2+3.3+4.4+5.5\right)}{3.4.\left(1+2.2+3.3+4.4+5.5\right)}\)

\(=\frac{1}{6}\)