Để chứng minh \(\frac{n+2011}{n+2012}\) là phân số tối giản => ( n+2011; n+2012 ) = 1

Gọi d là \(ƯCLN\left(n+2011;n+2012\right)\)

\(\Rightarrow\hept{\begin{cases}n+2011⋮d\\n+2012⋮d\end{cases}\Rightarrow\left(n+2012\right)-\left(n+2011\right)⋮d}\)

\(\Rightarrow1⋮d\Rightarrow d=1\)

\(\Rightarrow\frac{n+2011}{n+2012}\) là phân số tối giản.

gọi d là UCLN(n+2011,n+2012)

\(\Rightarrow\orbr{\begin{cases}n+2011⋮\\n+2012⋮\end{cases}}d\)

\(\Rightarrow\left(n+2012\right)-\left(n+2011\right)⋮d\)

\(\Rightarrow n+2012-n-2011⋮d\)

\(\Rightarrow1⋮d\Rightarrow d\inƯCLN\left(1\right)=\left\{\pm1\right\}\)

=> UCLN(N+2011,2012) = 1

=>\(\frac{2011}{2012}\)Là phân số tối giản

Chúc bạn học tốt !

\(\frac{1}{20\cdot23}+\frac{1}{23\cdot26}+...+\frac{1}{77\cdot80}\)

\(=\frac{1}{3}\left[\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+...+\frac{3}{77\cdot80}\right]\)

\(=\frac{1}{3}\left[\frac{1}{20}-\frac{1}{23}+...+\frac{1}{77}-\frac{1}{80}\right]\)

\(=\frac{1}{3}\left[\frac{1}{20}-\frac{1}{80}\right]\)

\(=\frac{1}{3}\left[\frac{4}{80}-\frac{1}{80}\right]\)

\(=\frac{1}{3}\cdot\frac{3}{80}=\frac{1}{1}\cdot\frac{1}{80}=\frac{1}{80}\)

Mà \(\frac{1}{80}< \frac{1}{9}\)nên \(\frac{1}{20\cdot23}+\frac{1}{23\cdot26}+...+\frac{1}{77\cdot80}< \frac{1}{9}\)

Vậy : ...

\(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\)

\(=\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)

\(=\frac{1}{3}.\frac{3}{80}\)

\(=\frac{1}{80}< \frac{1}{9}\)

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

\(\Rightarrow1\cdot1=xy\Rightarrow xy=1\)

Vì x, y là các số nguyên dương và xy = 1

\(\Rightarrow x=y=1\)

\(\frac{1}{x}=\frac{y}{2}-1\)

\(\Leftrightarrow\frac{1}{x}=\frac{y}{2}-\frac{2}{2}\)

\(\Leftrightarrow\frac{1}{x}=\frac{y-2}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}1=y-2\\x=2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=3\\x=2\end{cases}}\)

yếu tố gì của nó giảm

Ta có A=17^18+1/17^19+1 < 17^18+1+16/17^19+1+16 = 17^18+17/17^19+17 = 17(17^17+1/17^18+1)= B

Vậy A<B

\(A=\frac{17^{18}+1}{17^{19}+1}\)

Ta có :  \(17A=\frac{17(17^{18}+1)}{17^{19}+1}=\frac{17^{19}+17}{17^{19}+1}=\frac{17^{19}+1+16}{17^{19}+1}=1+\frac{17}{17^{19}+1}\)                    \((1)\)

\(B=\frac{17^{17}+1}{17^{18}+1}\)

Ta lại có : \(17B=\frac{17(17^{17}+1)}{17^{18}+1}=\frac{17^{18}+17}{17^{18}+1}=\frac{17^{18}+1+16}{17^{18}+1}=1+\frac{17}{17^{18}+1}\)        \((2)\)

Từ 1 và 2 suy ra : \(1+\frac{16}{17^{19}+1}< 1+\frac{16}{17^{18}+1}\)

Nên \(17A< 17B\)

Hay \(A< B\)

Vậy : \(A< B\)

\(3^8\cdot81\)

\(=3^8\cdot3^4\)

\(=3^{12}\)

Giúp tớ với mai mình phải nộp bài rồi. Cảm ơn

Ex1: It's getting light. Shall I turn...on.....the light to save electricity?

A) off.      B) on.          C) up.         D) in

Don't eat too much..salt... It's not good for you

A) vegetable     B) fish.         C) salt.          D) wine

Ex2: Complete these sentences with the words given.

1) Who/ be/ strong/ person/ your family?         Who is the strongest person in your family?

2) My/ future house/ be/ located/ the/ ocean.  My future house will be located in the ocean.

3) We/ have/ visit/ Cambodia/ last year             We had visited Cambodia last year.

4) Yesterday/ Nam/ go/ camping/ his classmates       Yesterday, Nam went camping with his classmates.

Ex 1 

A

C  hoặc D 

Ex 2 

1 Who are string person in your family

2.My future home will be located on the ocean 

3. We have visited Cambodia last year

4. Yesterday,Nam went camping with his classmates

Có thể có câu mk sai 

Bn giúp mk sủa lại 

Hok tốt ^^ 

============