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Ta có công thức tổng quát:
\(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)
\(a,A=\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}\\ =\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\dfrac{x-2}{5\left(x+3\right)}\\ =\dfrac{x-2}{15\left(x+3\right)}\)
Theo đề bài ta có:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)

=13/12x14/13x15/14x16/15x...x2006/2005x2007/2006x2008/2007
=2008/12
=502/3
A = 1\(\dfrac{1}{12}\) \(\times\) 1\(\dfrac{1}{13}\) \(\times\) 1\(\dfrac{1}{14}\) \(\times\) 1\(\dfrac{1}{15}\) \(\times\) ... \(\times\) 1\(\dfrac{1}{2005}\) \(\times\) 1\(\dfrac{1}{2006}\) \(\times\) 1\(\dfrac{1}{2007}\)
A = ( 1 + \(\dfrac{1}{12}\)) \(\times\) ( 1 + \(\dfrac{1}{13}\)) \(\times\) ( 1 + \(\dfrac{1}{14}\)) \(\times\)...\(\times\) ( 1 + \(\dfrac{1}{2006}\))\(\times\)(1+\(\dfrac{1}{2007}\))
A = \(\dfrac{13}{12}\) \(\times\) \(\dfrac{14}{13}\) \(\times\) \(\dfrac{15}{14}\) \(\times\) ...\(\times\) \(\dfrac{2007}{2006}\) \(\times\) \(\dfrac{2008}{2007}\)
A = \(\dfrac{13\times14\times15\times...\times2007}{13\times14\times15\times...\times2007}\) \(\times\) \(\dfrac{2008}{12}\)
A = 1 \(\times\) \(\dfrac{502}{3}\)
A = \(\dfrac{502}{3}\)

a; (5142 - 17 x 8 + 242 : 11) x (27 - 3 x 9)
= (5142 - 17 x 8 + 242 : 11) x (27 - 27)
= (5142 - 17 x 8 + 242 : 11) x 0
= 0
b;
(1 + \(\dfrac{1}{2}\)) \(\times\) (1 + \(\dfrac{1}{3}\)) \(\times\) ( 1 + \(\dfrac{1}{4}\)) \(\times\) ... \(\times\) (1 + \(\dfrac{1}{2010}\)) \(\times\)(1 + \(\dfrac{1}{2011}\))
= \(\dfrac{2+1}{2}\) \(\times\) \(\dfrac{3+1}{3}\) \(\times\) \(\dfrac{4+1}{4}\)\(\times\) ... \(\times\) \(\dfrac{2010+1}{2010}\)\(\times\) \(\dfrac{2011+1}{2011}\)
= \(\dfrac{3}{2}\)\(\times\)\(\dfrac{4}{3}\)\(\times\)\(\dfrac{5}{4}\)\(\times\)...\(\times\)\(\dfrac{2011}{2010}\)\(\times\)\(\dfrac{2012}{2011}\)
= \(\dfrac{2012}{2}\)
= 1006

\(71+52,5\times4=\frac{x+140}{x}+210\)
\(71+210=\frac{x+140}{x}+210\)
\(=>\frac{x+140}{x}=71\)
\(71=\frac{142}{2}\)\(\Rightarrow x=142-140=2\)

\(\dfrac{13+x}{20}\) = \(\dfrac{3}{4}\)
13 + \(x\) = 20 \(\times\) \(\dfrac{3}{4}\)
13 + \(x\) = 15
\(x\) = 15 - 13
\(x\) = 2
Cách khác :
\(\dfrac{13+x}{20}=\dfrac{3}{4}\)
\(\dfrac{13+x}{20}=\dfrac{15}{20}\)
\(13+x=15\)
\(x=15-13\)
\(x=2\)

3/4 * x + 1/2 * x -15 = 35
3/4 * x +1/2 * x = 35 + 15
3/4 * x +1/2 * x = 50
x * ( 3/4 + 1/2 ) = 50
x * 5/4 = 50
x = 50 : 5/4
x = 40
phan b mik ko nhap dc nen bn tu lm nha
a, \(\frac{3}{4}\times x+\frac{1}{2}\times x-15=35\)
\(x+x-15=\frac{3}{4}-\frac{1}{2}\)
\(x+x-15=\frac{2}{8}=\frac{1}{4}\)
\(x=35-15\)
\(x=20\)
Vậy \(x=\frac{1}{4}\)và \(x=20\)
b, Chịu

a; A = \(\dfrac{4026\times2014+4030}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2014+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-2013\times2+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-4026+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-2011\right)}{2013\times2016-2011}\)
A = 2

Ta có : \(\frac{1}{4}+\frac{1}{3}:\frac{1}{x}=\frac{11}{12}\)
\(\Rightarrow\frac{1}{3}:\frac{1}{x}=\frac{11}{12}-\frac{1}{4}\)
\(\frac{1}{3}:\frac{1}{x}=\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}:\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}\times\frac{3}{2}\)
\(\frac{1}{x}=\frac{1}{2}\)
=> x = 2
a) \(\frac{x\div3-16}{2}+21=38\)
\(\frac{x\div3-16}{2}=38+21\)
\(\frac{x\div3-16}{2}=59\)
\(x\div3-16=59.2\)
\(x\div3-16=118\)
\(x\div3=118+16\)
\(x\div3=134\)
\(x=134.3\)
\(x=402\)
b) \(\frac{1}{4}+\frac{1}{3}\div\frac{1}{x}=\frac{11}{12}\)
\(\frac{1}{3}\div\frac{1}{x}=\frac{11}{12}-\frac{1}{4}\)
\(\frac{1}{3}\div\frac{1}{x}=\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}\div\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{2}\)
Vậy x = ....

a, 7/5 : x + 3/2 = 16/3
7/5 : x = 16/3 - 3/2
7/5 : x = 23/6
x = 7/5 : 23/6
x = 42/115
b, x : 1/5 + 1/7 = 3/5 . 18/21
x : 1/5 + 1/7 = 18/35
x : 1/5 = 18/35 - 1/7
x : 1/5 = 13/35
x = 13/35 . 1/5
x = 13/175
c, x - 1 và 1/3 : 2 = 5/7
x - 4/3 : 2 = 5/7
x - 4/3 = 5/7 . 2
x - 4/3 = 10/7
x = 10/7 + 4/3
x = 58/21
d, x + 2 và 3/5 . 1/6 = 35/36
x + 13/5 . 1/6 = 35/36
x + 13/5 = 35/36 : 1/6
x + 13/5 = 35/6
x = 35/6 - 13/5
x = 97/30
e, ( x + 3/2 ) : 2 = 7/10 + 1/5
( x + 3/2 ) : 2 = 9/10
x + 3/2 = 9/10 . 2
x + 3/2 = 9/5
x = 9/5 - 3/2
x = 3/10

Câu 2.
=>x +4<11
=>x=0;1;2;3;4;5;6
=>x +4>9
=>x=6;7;8;9;...
=> x=6
\(\dfrac{21}{x}\cdot3=2\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{21}{x}=\dfrac{7}{9}\)
hay \(x=\dfrac{189}{7}\)