a) \(\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{6-1}{8}=\dfrac{5}{8}\)

b) \(\dfrac{2}{6}-\dfrac{5}{18}=\dfrac{6}{18}-\dfrac{5}{18}=\dfrac{6-5}{18}=\dfrac{1}{18}\)

c) \(\dfrac{2}{5}-\dfrac{3}{20}=\dfrac{8}{20}-\dfrac{3}{20}=\dfrac{8-3}{20}=\dfrac{5}{20}=\dfrac{1}{4}\)

1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)

\(\Leftrightarrow5x+20+12x-28=7x+2\)

\(\Leftrightarrow17x-7x=2+8=10\)

hay x=1

2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)

\(\Leftrightarrow6x+4-12x=-3x+3\)

\(\Leftrightarrow-6x+3x=3-4\)

hay \(x=\dfrac{1}{3}\)

3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)

\(\Leftrightarrow4x-12-x-2=6x-3\)

\(\Leftrightarrow3x-14-6x+3=0\)

\(\Leftrightarrow-3x=11\)

hay \(x=-\dfrac{11}{3}\)

4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)

\(\Leftrightarrow3x-6-8x-12=x+6\)

\(\Leftrightarrow-5x-x=6+18\)

hay x=-4

5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)

\(\Leftrightarrow6x-3+2x-6=-1\)

\(\Leftrightarrow8x=8\)

hay x=1

+) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)

\(\Rightarrow A=1-\dfrac{1}{2^{10}}=\dfrac{2^{10}-1}{2^{10}}\)

Vậy \(A=\dfrac{2^{10}-1}{2^{10}}\)

+) \(F=\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{1}{190}\)

\(\Rightarrow\dfrac{1}{2}F=\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{380}\)

\(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{19.20}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

\(=\dfrac{1}{5}-\dfrac{1}{20}=\dfrac{3}{20}\Rightarrow F=\dfrac{3}{20}:\dfrac{1}{2}=\dfrac{3}{10}\)

Vậy \(F=\dfrac{3}{10}\)

+) \(G=\dfrac{12}{84}+\dfrac{12}{210}+\dfrac{12}{390}+...+\dfrac{12}{2100}\)

\(=\dfrac{4}{28}+\dfrac{4}{70}+\dfrac{4}{130}+...+\dfrac{4}{700}=\dfrac{4}{4.7}+\dfrac{4}{7.10}+...+\dfrac{4}{25.28}\)

\(=\dfrac{4}{3}.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{25.28}\right)\)

\(=\dfrac{4}{3}.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(=\dfrac{4}{3}.\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{4}{3}.\dfrac{3}{14}=\dfrac{2}{7}\)

Vậy \(G=\dfrac{2}{7}\)

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)

\(A=1-\dfrac{1}{2^{10}}=\dfrac{1024-1}{1024}=\dfrac{1023}{1024}\)

\(F=\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{1}{190}\)

\(=\dfrac{2}{30}+\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{380}\)

\(=\dfrac{2}{5.6}+\dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{19.20}\)

\(=2\left(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{19.20}\right)\)

\(=2\left(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)

\(=2\left(\dfrac{1}{5}-\dfrac{1}{20}\right)=2.\dfrac{3}{20}=\dfrac{3}{10}\)

\(G=\dfrac{12}{84}+\dfrac{12}{210}+\dfrac{12}{390}+...+\dfrac{12}{2100}\)

\(=\dfrac{4}{28}+\dfrac{4}{70}+\dfrac{4}{130}+...+\dfrac{4}{700}\)

\(=\dfrac{4}{4.7}+\dfrac{4}{7.10}+\dfrac{4}{10.13}+...+\dfrac{4}{25.28}\)

\(=\dfrac{4}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)

\(=\dfrac{4}{3}.\dfrac{3}{14}=\dfrac{2}{7}\)

a, Theo bài ra ta có:

\(M=\dfrac{2007}{1}+1+\dfrac{2006}{2}+1+.......+\dfrac{2}{2006}+1+\dfrac{1}{2007}+1-2007\)

( Ta thêm 1 vào mỗi một số hạng trong M nên phải bớt đi 2017 vì có 2017 số hạng ) ;'

\(=>M=2008+\dfrac{2008}{2}+\dfrac{2008}{3}+......+\dfrac{2008}{2007}+\dfrac{2008}{2007}-2007\)

\(=>M=\dfrac{2008}{2}+\dfrac{2008}{3}+\dfrac{2008}{4}+.....+\dfrac{2008}{2006}+\dfrac{2008}{2007}+1\)

Ta thấy xuất hiện 2008 chung nên đặt ra ngoài ta có:

\(=>M=2008\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+....+\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}\right)\)

\(=>M:N=2008\)

Câu b đợi 1 chút nha.......

b, \(M=\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{31.33}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{31.33}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{31}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{33}\)

\(N=\dfrac{12}{11.13.15}+\dfrac{12}{13.15.17}+...+\dfrac{12}{31.33.35}\)

\(=3\left(\dfrac{4}{11.13.15}+\dfrac{4}{13.15.17}+...+\dfrac{4}{31.33.35}\right)\)

\(=3\left(\dfrac{1}{11.13}-\dfrac{1}{13.15}+\dfrac{1}{13.15}-\dfrac{1}{15.17}+...+\dfrac{1}{31.33}-\dfrac{1}{33.35}\right)\)

\(=3\left(\dfrac{1}{11.13}-\dfrac{1}{33.35}\right)\)

\(=\dfrac{92}{5005}\)

\(\Rightarrow M:N=\dfrac{1}{33}:\dfrac{92}{5005}=\dfrac{455}{276}\)

Vậy...

\(A=\dfrac{636363\cdot37-373737\cdot63}{1+2+3+...+2006}\)

\(=\dfrac{37^2\cdot3^3\cdot7^2\cdot13-37^2\cdot3^3\cdot7^2\cdot13}{\left(2006+1\right)\cdot1003}\)

=0

Em xin lỗi, p.số cuối ở số 1 ở ngoài căn ạ, em đánh lộn: 

Áp dụng hằng đẳng thức: \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\) nếu \(a+b+c=0\) là ra

ĐK: \(n>0\)

Ta có: \(\left(1+\dfrac{1}{n}-\dfrac{1}{n+1}\right)^2\)

\(=1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}+2\left(\dfrac{1}{n}-\dfrac{1}{n+1}-\dfrac{1}{n\left(n+1\right)}\right)\)

\(=1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}+2.\dfrac{n+1-n-1}{n\left(n+1\right)}\)

\(=1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}+0=1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}\)

Áp dụng:

\(\sqrt{1+\dfrac{1}{11^2}+\dfrac{1}{12^2}}+\sqrt{1+\dfrac{1}{12^2}+\dfrac{1}{13^2}}+...+\sqrt{1+\dfrac{1}{\left(n+1\right)^2}+\dfrac{1}{\left(n+2\right)^2}}\)

\(=\sqrt{\left(1+\dfrac{1}{11}-\dfrac{1}{12}\right)^2}+\sqrt{\left(1+\dfrac{1}{12}-\dfrac{1}{13}\right)^2}+...+\sqrt{\left(1+\dfrac{1}{n+1}-\dfrac{1}{n+2}\right)^2}\)

\(=\left|1+\dfrac{1}{11}-\dfrac{1}{12}\right|+\left|1+\dfrac{1}{12}-\dfrac{1}{13}\right|+...+\left|1+\dfrac{1}{n+1}-\dfrac{1}{n+2}\right|\)

\(=1+\dfrac{1}{11}-\dfrac{1}{12}+1+\dfrac{1}{12}-\dfrac{1}{13}+...+1+\dfrac{1}{n+1}-\dfrac{1}{n+2}\)

\(=\left(n-9\right)+\dfrac{1}{11}-\dfrac{1}{n+2}=n-\dfrac{1}{n+2}-\dfrac{98}{11}\)

`a)1/2+[-1]/[-3]-5/12 < 2x < 12/[-31]+136/31`

`186/372+124/372-155/372 < [744x]/372 < [-144]/372+1632/372`

`186+124-155 < 744x < -144+1632`

`155 < 744x < 1488`

`155:744 < 744x:744 < 1488:744`

`5/24 < x < 2`

Vậy `5/24 < x < 2`

__________________________________________________

`b)[-2]/5 < x/15 < 1/6`

`[-12]/30 < [2x]/30 < 5/30`

`-12 < 2x < 5`

`-12:2 < 2x:2 < 5:2`

`-6 < x < 5/2`

Vậy `-6 < x < 5/2`

Giải:

a) x - \(\dfrac{9}{25}\)= \(\dfrac{16}{25}\)

x = \(\dfrac{16}{25}\)+\(\dfrac{9}{25}\)
x = \(\dfrac{25}{25}\)

x = 1

b) \(\dfrac{-12}{30}\)<\(\dfrac{x}{30}\)<\(\dfrac{5}{30}\)

=> x có thể bằng \(\dfrac{-11}{30}\) đến \(\dfrac{4}{30}\)
=> x bằng -5; -4; -3; -2; -1;0;1;2

a) Ta có: \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)

\(\Leftrightarrow\dfrac{2\left(x+5\right)}{6\left(x-2\right)}-\dfrac{3\left(x-2\right)}{6\left(x-2\right)}=\dfrac{3\left(2x-3\right)}{6\left(x-2\right)}\)

Suy ra: \(2x+5-3x+6=6x-9\)

\(\Leftrightarrow-x+11-6x+9=0\)

\(\Leftrightarrow20-7x=0\)

\(\Leftrightarrow7x=20\)

hay \(x=\dfrac{20}{7}\)(thỏa ĐK)

Vậy: \(S=\left\{\dfrac{20}{7}\right\}\)