`1/8+1/24+1/48+1/80+1/120`

`=1/[2xx4]+1/[4xx6]+1/[6xx8]+1/[8xx10]+1/[10xx12]`

`=1/2xx(2/[2xx4]+2/[4xx6]+2/[6xx8]+2/[8xx10]+2/[10xx12])`

`=1/2xx(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12)`

`=1/2xx(1/2-1/12)`

`=1/2xx(6/12-1/12)`

`=1/2xx5/12=5/24`

\(\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)

=\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{10.12}\)

=\(\dfrac{1}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{10.12}\right)\)

=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\)

=\(\dfrac{1}{2}.\dfrac{5}{12}\)

=\(\dfrac{5}{24}\)

Dấu chấm(.)là nhân.

   \(\dfrac{1}{8}\) + \(\dfrac{1}{24}\)+ \(\dfrac{1}{48}\)+ \(\dfrac{1}{80}\)+ \(\dfrac{1}{120}\)

 = \(\dfrac{1}{2}\) X (\(\dfrac{2}{2\times4}\)+ \(\dfrac{2}{4\times6}\)+\(\dfrac{2}{6\times8}\)+\(\dfrac{2}{8\times10}\)+ \(\dfrac{2}{10\times12}\))

= \(\dfrac{1}{2}\) x ( \(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)-\(\dfrac{1}{12}\))

= \(\dfrac{1}{2}\)x (\(\dfrac{1}{2}\)-\(\dfrac{1}{12}\))

= \(\dfrac{1}{2}\) X \(\dfrac{6-1}{12}\)

= \(\dfrac{1}{2}\)x \(\dfrac{5}{12}\)

= \(\dfrac{5}{24}\)

=\(1+\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{10\cdot12}\)

\(=1+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{12}\right)\)

\(=1+\dfrac{1}{2}\cdot\dfrac{5}{12}=1+\dfrac{5}{24}=\dfrac{29}{24}\)

==1+12(12−14+14−16+...+110−112)=1+12(12−14+14−16+...+110−112)

\(=\dfrac{1}{8}\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)\)

\(=\dfrac{1}{8}\cdot2\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)

\(=\dfrac{1}{4}\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{6}{7}=\dfrac{3}{14}\)

\(1,=2\sqrt{3}-3\sqrt{3}+4\sqrt{3}=3\sqrt{3}\\ 2,=\left(2\sqrt{6}+2\sqrt{5}-4\sqrt{5}\right):5=\dfrac{2\sqrt{6}}{5}-\dfrac{2\sqrt{5}}{5}\\ 3,=6\sqrt{3}-\dfrac{4\sqrt{3}}{3}-4\sqrt{3}-\dfrac{5\sqrt{3}}{3}=2\sqrt{3}-\dfrac{9\sqrt{3}}{3}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\\ 4,Sửa:\dfrac{1}{\sqrt{5}-\sqrt{3}}-\dfrac{1}{\sqrt{5}+\sqrt{3}}\\ =\dfrac{\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)

1) \(=2\sqrt{3}-3\sqrt{3}+4\sqrt{3}=3\sqrt{3}\)

2) \(=\left(2\sqrt{6}+2\sqrt{5}-4\sqrt{5}\right)=\dfrac{2\sqrt{6}}{5}+\dfrac{2\sqrt{5}}{5}-\dfrac{4\sqrt{5}}{5}\)

3) \(=6\sqrt{3}-\dfrac{4\sqrt{3}}{3}-4\sqrt{3}-\dfrac{5\sqrt{3}}{3}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\)

4) \(=\dfrac{\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{5-3}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)

133/99

quy đòng r tính nha ra \(\dfrac{199}{33}\)

A= 1/3+1/6+1/12+1/24+1/48+1/96

  = (1/3+1/6)+(1/12+1/24)+(1/48+1/96)

  = (2/6+1/6)+(2/24+1/24)+(2/96+1/96)

  = 1/2+1/8+1/32

  = 16/32+4/32+1/32
  = 21/32

Vậy A=21/32

Giải:

A=1/3+1/6+1/12+1/24+1/48+1/96

A=1/3+(1/2.3+1/3.4)+(1/4.6+1/6.8)+1/96

A=1/3+(1/2-1/3+1/3-1/4)+[1/2.(2/4.6+2/6.8)]+1/96

A=1/3+(1/2-1/4)+[1/2.(1/4-1/6+1/6-1/8)]+1/96

A=1/3+1/4+[1/2.(1/4-1/8)]+1/96

A=1/3+1/4+[1/2.1/8]+1/96

A=1/3+1/4+1/16+1/96

A=7/12+7/96

A=21/32

câu b bài 2:

\(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)

\(=\dfrac{1}{5}\)

câu a bài 2:

\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)

\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)

1. \(\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).0\)

\(=0\)

Bài 1: 

Ta có: \(A=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)

=0

bỏ số 1 ở đầu thì giải dc á, còn có số 1 thì chịu

\(\dfrac{1}{x+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)

\(\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+8\right)}=\dfrac{4}{105}\)

\(\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)

\(\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+8}=\dfrac{8}{105}\)

\(\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)

\(\dfrac{x+8-x}{x\left(x+8\right)}=\dfrac{8}{105}\)

\(\dfrac{8}{x.\left(x+8\right)}=\dfrac{8}{105}\)

\(\Rightarrow x\left(x+8\right)=105\)

\(x^2+8x-105=0\)

\(\left(x-7\right)\left(x+15\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)

không có số 1 ở đầu đâu.Mong mọi người giải giúp mk nhé!