\(\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}\)
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23 tháng 8 2018

=1/8-1/5+1/7*3/8-3/5+3/7 + 1/2+1/3-1/5*3/4+1/2-3/10

=19/280*57/280+19/30*19/20

=19/280.280/57+19/30.20/19

=1/1.1/3+1/3.2/1

=1/3+2/3=3/3

=1

23 tháng 8 2018

* : chia

. : nhân

18 tháng 6 2018

1)

\(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\)
\(=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{6}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\)
\(=\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\)
\(=\dfrac{3}{-5}+\dfrac{3}{5}\)
\(=-\dfrac{3}{5}+\dfrac{3}{5}\)
\(=0\)

\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}\)

\(=\dfrac{1}{2}\cdot\dfrac{3}{4}=\dfrac{3}{8}\)

a: \(=\dfrac{2}{3}\left(\dfrac{3}{60\cdot63}+\dfrac{3}{63\cdot66}+...+\dfrac{3}{117\cdot120}\right)+\dfrac{2}{2006}\)

\(=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{63}+...+\dfrac{1}{117}-\dfrac{1}{120}\right)+\dfrac{2}{2006}\)

\(=\dfrac{2}{3}\cdot\dfrac{1}{120}+\dfrac{1}{2003}=\dfrac{1}{180}+\dfrac{1}{2003}=\dfrac{2183}{180\cdot2003}\)

b: \(=\dfrac{5}{4}\left(\dfrac{4}{40\cdot44}+\dfrac{4}{44\cdot48}+...+\dfrac{4}{76\cdot80}\right)+\dfrac{5}{2006}\)

\(=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{80}\right)+\dfrac{5}{2006}\)

\(=\dfrac{5}{4}\cdot\dfrac{1}{80}+\dfrac{5}{2006}=\dfrac{1}{64}+\dfrac{5}{2006}=\dfrac{1163}{64192}\)

c: \(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}+\dfrac{3}{17\cdot20}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{3}\cdot\dfrac{9}{20}=\dfrac{3}{20}\)

 

25 tháng 12 2017

Tính 1 câu thoy nhé !

\(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)

= \(\dfrac{3}{7}.\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)\)

=\(\dfrac{3}{7}.-14=-6\)

14 tháng 12 2018

\(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{4031}{2015^2.2016^2}\)

\(A=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{2016^2-2015^2}{2015^2.2016^2}\)

\(A=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{2015^2}-\dfrac{1}{2016^2}\)

\(A=1-\dfrac{1}{2016^2}< 1\left(đpcm\right)\)

20 tháng 9 2017

Mấy bài dễ tự làm nhé:D

1)

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\\\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\end{matrix}\right.\)

Ta có điều phải chứng minh

\(\left\{{}\begin{matrix}\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\\\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\end{matrix}\right.\)

Ta có điều phải chứng minh

6 tháng 12 2017

a. \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)

\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)

\(=1+1+0,5\)

\(=2,5\)

b. \(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)

\(=\dfrac{3}{7}.\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)\)

\(=\dfrac{3}{7}.\left(-14\right)=-6\)

c. \(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(\dfrac{-5}{7}\right)\)

\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)\)

\(=-10:\left(-\dfrac{5}{7}\right)\)

\(=14\)

d. \(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\dfrac{-5}{21}:\dfrac{4}{5}+\dfrac{5}{21}:\dfrac{4}{5}\)

\(=\left(\dfrac{-5}{7}+\dfrac{5}{7}\right):\dfrac{4}{5}\)

\(=0:\dfrac{4}{5}\)

\(=0\)

6 tháng 12 2017

a,

\(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)

\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)

\(=1+1-0,5=1,5\)

b,

\(\dfrac{3}{7}\cdot19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)

\(=\dfrac{3}{7}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{3}{7}.\left(-14\right)=-6\)

c,

\(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)\)

\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)=-10:\left(-\dfrac{5}{7}\right)=14\)

d,

\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\left(-\dfrac{2}{3}+\dfrac{3}{7}+\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\left[\left(-\dfrac{2}{3}+\dfrac{-1}{3}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\right]:\dfrac{4}{5}\)

\(=\left(-1+1\right):\dfrac{4}{5}=0:\dfrac{4}{5}=0\)

a: \(=\left(1+\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)

\(=1+1+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)

b: \(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)

\(=\dfrac{31}{25}:\dfrac{-29}{25}=\dfrac{-31}{29}\)

c: \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)

=1/4+3/4

=1