A=\(\frac{1.2.3.4...8.9}{2.3.4.5...9.10}\)

A=\(\frac{1}{10}\)

mình làm đc 1 câu thôi. Bạn thông cảm nhé

Rút gọn :

\(\frac{-63}{81}=\frac{-7\cdot9}{9\cdot9}=\frac{-7}{9}\)

\(\frac{-5\cdot6}{9\cdot35}=\frac{-5\cdot2\cdot3}{3\cdot3\cdot5\cdot7}=\frac{-2}{3\cdot7}=\frac{-2}{21}\)

\(\frac{7\cdot2+8}{2\cdot14\cdot5}=\frac{14+8}{2\cdot2\cdot7\cdot5}=\frac{22}{2\cdot2\cdot7\cdot5}=\frac{2\cdot11}{2\cdot2\cdot7\cdot5}=\frac{11}{2\cdot7\cdot5}=\frac{11}{70}\)

\(\frac{-63}{81}=\frac{-7}{9}\)

\(\frac{-5.6}{9.35}=\frac{-5.2.3}{3.3..5.7}=\frac{-2}{3.7}=\frac{-2}{21}\)

\(\frac{7.2+8}{2.14.5}=\frac{22}{2.2.7.5}=\frac{2.11}{2.2.7.5}=\frac{11}{70}\)

a)\(=\dfrac{211}{180}\)

b)\(=\dfrac{5}{39}\)

c)=\(=-\dfrac{65}{168}\)

\(E=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}...\frac{9^2}{8.10}=\frac{\left(2.3.4...9\right)^2}{1.2.\left(3.4...8\right)^2.9.10}=\frac{2^2.9^2}{1.2.9.10}=\frac{18}{10}=\frac{9}{5}\)

\(\frac{x-1}{2018}+\frac{x-7}{503}=\frac{x-3}{1008}+\frac{x-9}{670}\)

\(\Leftrightarrow\frac{x-1}{2018}-1+\frac{x-7}{503}-4=\frac{x-3}{1008}-2+\frac{x-9}{670}-3\)

\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{2018}+\frac{1}{503}-\frac{1}{1008}-\frac{1}{670}\right)=0\)

\(\Rightarrow x=2019\)

#CBHT

Đặt A =1/2+1/4+1/8+...+1/1024

2A= 1+1/2+1/4+...+1/512

A= 1-1/1024

=>A<1hay ...

\(\frac{x}{63}=\frac{-20}{63}\)

\(\Rightarrow x=-20\)

\(\frac{x}{63}=\frac{-20}{63}\)

=>x=-20

kết quả bằng 9/5 đó bạn

giải ra các cách được hông bạn

a. \(\left(x+8\right)⋮\left(x+4\right)\)

\(\Rightarrow\left(x+4\right)+4⋮\left(x+4\right)\)

Mà \(\left(x+4\right)⋮\left(x+4\right)\)

\(\Rightarrow4⋮\left(x+4\right)\)

\(\Rightarrow x+4\in\text{Ư} \left(4\right)=\left\{1;2;4\right\}\)

Ta có 3 trường hợp :

TH1 : \(x+4=1\Rightarrow x\notin N\) ( Loại )

TH2 : \(x+4=2\Rightarrow x\notin N\)(Loại )

TH3 : \(x+4=4\Rightarrow x=0\)

Vậy x = 0

a,Vì : \(x+8⋮x+2\)

Mà : \(x+2⋮x+2\)

\(\Rightarrow\left(x+8\right)-\left(x+2\right)⋮x+2\Rightarrow x+8-x-2⋮x+2\)

\(\Rightarrow6⋮x+2\Rightarrow x+2\inƯ\left(6\right)\)

Mà : \(Ư\left(6\right)=\left\{1;2;3;6\right\}\) ; \(x+2\ge2\Rightarrow x+2\in\left\{2;3;6\right\}\)

\(\Rightarrow x\in\left\{0;1;4\right\}\)

Vậy ...

b,Ta có : \(2y+7⋮y-1\) ; \(y-1⋮y-1\Rightarrow2\left(y-1\right)⋮y-1\Rightarrow2y-2⋮y-1\)

\(\Rightarrow\left(2y+7\right)-\left(2y-2\right)⋮y-1\Rightarrow2y+7-2y+2⋮y-1\)

\(\Rightarrow9⋮y-1\Rightarrow y-1\in\left\{1;3;9\right\}\Rightarrow y\in\left\{2;4;10\right\}\)

Vậy ...

c, Vì : \(x\in N\Rightarrow x-5\in N\)

\(y\in N\Rightarrow y+3\in N\left(y+3\ge3\right)\)

\(\Rightarrow x-5,y+3\inƯ\left(7\right)\)

Mà : \(Ư\left(7\right)=\left\{1;7\right\};y+3\ge3\)

\(\Rightarrow x-5=1\Rightarrow x=6;y+3=7\Rightarrow y=4\)

Vậy ...