\((x-3)(2x-7)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x-7=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{7}{2}\end{cases}}\)

Vậy \(x\in\left\{3;\frac{7}{2}\right\}\)

\(\left(x-3\right)\left(2x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=\frac{7}{2}\end{cases}}\)

Vậy ...

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 : \(A< B\)

\(3^8\cdot81\)

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

\(=3^{12}\)