4.

tử của A= 4(1.2 + 2.4+...+8.10) 

mẫu của A= 1.2 + 2.4+...+8.10 nên A=4

\(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{x\left(x+2\right)}=\dfrac{2011}{2013}\)

\(\Leftrightarrow2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{2011}{2013}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x-2}=\dfrac{2011}{4026}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x-2}=\dfrac{2011}{4026}\)\(\Leftrightarrow\dfrac{1}{x-2}=\dfrac{1}{2013}\)

\(\Rightarrow x-2=2013\Rightarrow x=2015\)

\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}\)= \(\frac{1}{8}\)

\(\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{\left(2x-2\right).2x}\right)\)= \(\frac{1}{8}\)

\(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2x}\right)\)= \(\frac{1}{8}\)

=> \(\frac{1}{2}-\frac{1}{2x}\)= \(\frac{1}{4}\)

=> 1/2x = 1/4

=> 2x = 4

x = 4 : 2

x = 2

\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right)2x}=\frac{1}{8}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}\right)=\frac{1}{8}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{8}.2=\frac{1}{4}\)

\(\Rightarrow\frac{1}{2x}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\)

\(\Rightarrow2x=4\Leftrightarrow x=2\)

1/3*x+2/5*(x-1)=4

1/3*x+2/5*x-2/5=4

1/3*x+2/5*x=4+2/5

(1/3+2/5)*x=4 2/5

11/15*x=22/5

         x=22/5/11/15

         x=22/5*11/15

         x=242/75

ủa vậy ư

a) ( x - 3) : 25 = 28

x - 3 = 28 x 25

x - 3 = 700

x = 700 + 3 = 703

b) ( 8 - x : 2 ) : 4 = 2

8 - x : 2 = 2 x 4

8 - x : 2 = 8

x : 2 = 8 - 8 = 0

=> x = 0

Bài làm :

\(a,\left(x-2\right):25=28\)

\(x-2=28.25\)

\(x-2=700\)

\(x=700+2\)

\(x=702\)

Vậy x = 702 .

\(b,\left(8-x:2\right):4=2\)

\(8-x:2=8\)

\(x:2=8-8\)

\(x:2=0\)

\(x=0\)

Vậy x = 0.

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