b: \(\dfrac{121212}{424242}=\dfrac{12}{42}=\dfrac{2}{7}\)

c: \(\dfrac{187187187}{221221221}=\dfrac{187}{221}=\dfrac{11}{13}\)

a) Ta có : 
\(\underbrace{99...9}5 = 5 \times 1 \underbrace{99...9}\)
100 chữ số 9            100 chữ số 9
Vậy ...

121212/42424 = 2/7

187187187/22122122111 = 11/13

3.7,13,37,39-10101/505050+70707 = chó nào mà giải dc

chính cậu ko giải dc nữa còn đố

nếu mjnk trả loi dc thjj sao

a.12/42

b.2.000000001

a)121212/424242=12/42=2/7

b)19999999999/99999999995=1/5

\(a,\)\(\frac{121212}{424242}\)\(=\)\(\frac{121212:60606}{424242:60606}\)\(=\)\(\frac{2}{7}\)

\(b,\)\(\frac{19999999999}{99999999995}\)\(=\)\(\frac{19999999999:19999999999}{99999999995:19999999999}\)\(=\)\(\frac{1}{5}\)

k cho mình nha chúc bạn học tốt

ai k cho mình mình k lai

Bài 1 .

a) Gọi d \(\in\)ƯC ( n + 1 , 2n + 3 ) . Ta có :

2n + 3 - 2( n + 1 ) \(⋮\)cho d

\(\Rightarrow\)1 chia hết cho d => d = + , - 1

b ) Gọi d \(\in\)ƯC ( 2n + 3 , 4n + 8 ) . Ta có :

4n + 8 - 2( 2n + 3 ) \(⋮\)cho d

\(\Rightarrow\)2 chia hết cho d . Do đó d là Ư của số lẻ 2n + 3 nên d = + , - 1

c ) Xét buểu thức 5( 3n + 2 ) - 3( 5n + 3 ).

\(\frac{2}{3}\)

\(\frac{3.7.13.37.19-10101}{505050+70707}\)=\(\frac{10101.39-10101}{10101.50+10101.7}\)=\(\frac{10101.\left(39-1\right)}{10101.\left(50+7\right)}\)=\(\frac{38}{57}\)=\(\frac{2}{3}\)

\(\frac{3.7.13.37.39-10101}{505050+70707}\)

\(=\frac{\left(3.7.13.37\right).39-10101}{50.10101+7.10101}\)

\(=\frac{10101.39-10101.1}{\left(50+7\right).10101}\)

\(=\frac{\left(39-1\right).10101}{57.10101}\)

\(=\frac{38.10101}{57.10101}\)

\(=\frac{38}{57}=\frac{38:19}{57:19}=\frac{2}{3}\)

a, \(\frac{n+2}{n+3}\)

Gọi \(d=ƯCLN\left(n+2,n+3\right)\)

\(\Rightarrow\hept{\begin{cases}n+2⋮d\\n+3⋮d\end{cases}}\)

\(\Rightarrow\left(n+3\right)-\left(n+2\right)⋮d\)

\(\Rightarrow1⋮d\Rightarrow d=1\)

Vậy phân số \(\frac{n+2}{n+3}\)là p/số tối giản

b, \(\frac{n+1}{2n+3}\)

Gọi \(d=ƯCLN\left(n+1,2n+3\right)\)

\(\Rightarrow\hept{\begin{cases}n+1⋮d\\2n+3⋮d\end{cases}\Rightarrow}\hept{\begin{cases}2\left(n+1\right)⋮d\\2n+3⋮d\end{cases}\Rightarrow}\hept{\begin{cases}2n+2⋮d\\2n+3⋮d\end{cases}}\)

\(\Rightarrow\left(2n+3\right)-\left(2n+2\right)⋮d\)

\(\Rightarrow1⋮d\Rightarrow d=1\)

Vậy...

c, \(\frac{2n+3}{4n+8}\)

Gọi \(d=ƯCLN\left(2n+3,4n+8\right)\)

\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+8⋮d\end{cases}\Rightarrow}\hept{\begin{cases}2\left(2n+3\right)⋮d\\4n+8⋮d\end{cases}\Rightarrow}\hept{\begin{cases}4n+6⋮d\\4n+8⋮d\end{cases}}\)

\(\Rightarrow\left(4n+8\right)-\left(4n+6\right)⋮d\)

\(\Rightarrow2⋮d\)

\(\Rightarrow d\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Nhưng 2n + 3 là số lẻ \(\Rightarrow\)d cũng là số lẻ

\(\Rightarrow d=1\)

Vậy....