Question

Tính giá trị của : \(\dfrac{1}{3\text{x}4}\)+\(\dfrac{1}{4\text{x}5}\)+\(\dfrac{1}{5\text{x}6}\)+...+\(\dfrac{1}{38\text{x}39}\)

gấp ạ

 

Answer 1

\(\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{38\times39}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...\dfrac{1}{38}-\dfrac{1}{39}\)

\(=\dfrac{1}{3}-\dfrac{1}{39}\)

\(=\dfrac{13}{39}-\dfrac{1}{39}\)

\(=\dfrac{12}{39}=\dfrac{4}{13}\)

Answer 2

\(\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+....+\dfrac{1}{38\times39}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{38}-\dfrac{1}{39}\)

\(=\dfrac{1}{3}-\dfrac{1}{39}\)

\(=\dfrac{13}{39}-\dfrac{1}{39}=\dfrac{12}{39}=\dfrac{4}{13}\)

Answer 3

`1/(3xx4)+1/(4xx5)+1/(5xx6)+...+1/(38xx39)`

`=1/3-1/4+1/4-1/5+1/5-1/6+...+1/38-1/39`

`=1/3-1/39`

`=4/13`

Answer 4

\(\dfrac{1}{3\text{x}4}+\dfrac{1}{4\text{x}5}+\dfrac{1}{5\text{x}6}+...+\dfrac{1}{38\text{x}39}\\ \\ \\ \\ \\ =\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{38}-\dfrac{1}{39}\\ \\ \\ \\ \\ =\dfrac{1}{3}-\dfrac{1}{39}=\dfrac{13}{39}-\dfrac{1}{39}=\dfrac{12}{39}\)

Answer 5

\(\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{38\times39}\)

= \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{38}-\dfrac{1}{39}\)

= \(\dfrac{1}{3}-\dfrac{1}{39}=\dfrac{4}{13}\)