Theo đề ta có \(\left(x+\frac{1}{y}\right)\in Z\) và \(\left(y+\frac{1}{x}\right)\in Z\)\(\Rightarrow\)\(\left(x+\frac{1}{y}\right)\left(y+\frac{1}{x}\right)\in Z\)

hay \(\left(xy+\frac{1}{xy}+2\right)\in Z\)\(\Rightarrow\)\(\left(xy+\frac{1}{xy}\right)\in Z\)

Suy ra \(\left(xy+\frac{1}{xy}\right)^2\in Z\)\(\Rightarrow\)\(\left(x^2y^2+\frac{1}{x^2y^2}+2\right)\in Z\)\(\Rightarrow\)\(\left(x^2y^2+\frac{1}{x^2y^2}\right)\in Z\)

Vậy \(x^2y^2+\frac{1}{x^2y^2}\) là số nguyên (đpcm).

\(\left(x+\frac{1}{y}\right)\left(y+\frac{1}{x}\right)=xy+2+\frac{1}{xy}\)

vì 2 nguyên nên \(xy+\frac{1}{xy}\)nguyên

\(\left(xy+\frac{1}{xy}\right)^2=x^2y^2+\frac{1}{x^2y^2}+2\)

nen \(x^2y^2+\frac{1}{x^2y^2}\)nguyên

abc=100a+10b+c=n²-1(*)

cba=100c+10b+a=n²-4n+4(**)

(*)-(**)=99(a-c)=4n+5

=> 4n-5 chia hết cho 99

Mà \(100\le abc\le999\)

=> \(100\le n^2-1\le999\)

<=> \(101\le n^2\le1000\)=\(11< 31\)=\(39\le4n-5\le199\)

Vì  4n+5 chia hết cho 99 

Nên 4n-5=99

4n=99+5

4n=104

n=104:4

n=26

Vậy abc=675

bạn ơi giúp mk giải nốt bài 2 đc ko ? cảm ơn bạn rất rất nhìu

a) \(\frac{1}{x}+\frac{1}{y}=2\Leftrightarrow\frac{x+y}{xy}=2\)

\(\Leftrightarrow x+y=2xy\Leftrightarrow4xy=2x+2y\)

\(\Leftrightarrow4xy-2x-2y=0\Leftrightarrow2x\left(2y-1\right)-\left(2y-1\right)=1\)

\(\Leftrightarrow\left(2x-1\right)\left(2y-1\right)=1=1.1=\left(-1\right).\left(-1\right)\)

\(TH1:\hept{\begin{cases}2x-1=1\\2y-1=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}\)

\(TH1:\hept{\begin{cases}2x-1=-1\\2y-1=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=0\end{cases}}\left(L\right)\)

Vậy x = y = 1

b) A là số chính phương nên ta đặt \(n^2+2n+8=a^2\)

\(\Leftrightarrow\left(n+1\right)^2+7=a^2\)

\(\Leftrightarrow a^2-\left(n+1\right)^2=7\)

\(\Leftrightarrow\left(a-n-1\right)\left(a+n+1\right)=7=1.7=7.1\)

\(=\left(-1\right).\left(-7\right)=\left(-7\right).\left(-1\right)\)

Lập bảng:

Mà n là số tự nhiên nên n = 2.

Vì \(x+\frac{1}{y}\in Z;y+\frac{1}{x}\in Z\)nên \(\left(x+\frac{1}{y}\right)\left(y+\frac{1}{x}\right)\in Z\)

=>\(xy+\frac{1}{xy}\in Z\)

=>\(\left(xy+\frac{1}{xy}\right)^3\)

=>\(x^3y^3+\frac{1}{x^3y^3}+3\left(xy+\frac{1}{xy}\right)\)\(\in Z\)

=>ĐPCM