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

Với mọi n thuộc N * ta có :

\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}}=\sqrt{\frac{n^4+2n^3+n^2+2n^2+2n+1}{n^2\left(n+1\right)^2}}\)

\(=\sqrt{\frac{n^4+2n^3+3n^2+2n+1}{n^2\left(n+1\right)^2}}=\sqrt{\frac{n^4+n^2+1+2n^3+2n+2n^2}{n^2\left(n+1\right)^2}}\)

\(=\sqrt{\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}}=\frac{n^2+n+1}{n\left(n+1\right)}=1+\frac{1}{n\left(n+1\right)}=1+\frac{1}{n}-\frac{1}{n+1}\)

Áp dụng vào ta được : 

\(A=\left(1+1-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+....+\left(1+\frac{1}{2011}-\frac{1}{2012}\right)\)

\(=2012-\frac{1}{2012}=\frac{2012^2-1}{2012}\)

 ai biết làm chỉ cho mik công thức với :(((

\(\frac{x^2}{y^2}+\frac{y^2}{x^2}-\frac{x}{y}-\frac{y}{x}\)

\(=\frac{x^2-xy}{y^2}+\frac{y^2-xy}{x^2}\)

\(=\frac{x^4-x^3y+y^4-xy^3}{x^2y^2}\)

\(=\frac{x^3\left(x-y\right)-y^3\left(x-y\right)}{x^2y^2}\)

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

2) Có: \(x^3+y^3=\sqrt{\left(x.x^2+y.y^2\right)^2}\le\sqrt{\left(x^2+y^2\right)\left(x^4+y^4\right)}\)

And: \(\sqrt{x^3y^3}=\left(\sqrt{xy}\right)^6\le\left(\frac{x+y}{2}\right)^6=1\)

\(\Rightarrow\)\(x^3y^3\left(x^3+y^3\right)\le\sqrt{x^3y^3}\sqrt{x^3y^3\left(x^2+y^2\right)\left(x^4+y^4\right)}=\sqrt{xy\left(x^2+y^2\right).x^2y^2\left(x^4+y^4\right)}\)

Theo bài 1 thì \(xy\left(x^2+y^2\right)\le2\) do đó theo cách đặt \(x^2=a;y^2=b\) ta cũng có: \(x^2y^2\left(x^4+y^4\right)=ab\left(a^2+b^2\right)\le2\)

Do đó: \(x^3y^3\left(x^3+y^3\right)\le\sqrt{2.2}=2\) ( đpcm ) 

\(VT=\frac{x^4}{x^4+3xyzt}+\frac{y^4}{y^4+3xyzt}+\frac{z^4}{z^4+3xyzt}\ge\frac{\left(x^2+y^2+z^2+t^2\right)^2}{x^4+y^4+z^4+t^4+12xyzt}\)

Có: \(4abcd=4\sqrt{a^2b^2.c^2d^2}\le2\left(a^2b^2+c^2d^2\right)\)

Tương tự, ta cũng có: 

\(4abcd\le2\left(a^2c^2+b^2d^2\right)\)

\(4abcd\le2\left(d^2a^2+b^2c^2\right)\)

\(\Rightarrow\)\(VT\ge\frac{\left(x^2+y^2+z^2+t^2\right)^2}{x^4+y^4+z^4+t^4+2\left(xy+yz+zt+tx+yz+zt\right)}=1\) ( đpcm ) 

Với \(y=1\Rightarrow\dfrac{x^2+x+1}{x+1}\in Z\Rightarrow\dfrac{1}{x+1}\in Z\Rightarrow\) ko tồn tại x nguyên dương thỏa mãn (loại)

Với \(y>1\):

Đặt \(\dfrac{x^2+x+1}{xy+1}=k\Rightarrow x^2-\left(ky-1\right)x+1-k=0\)

\(\Delta=\left(ky-1\right)^2+4\left(k-1\right)\) là số chính phương

Ta có: \(k\ge1\Rightarrow\left(ky-1\right)^2+4\left(k-1\right)\le\left(ky-1\right)^2\)

Đồng thời \(y>1\Rightarrow y\ge2\Rightarrow2ky\ge4k>3\)

\(\Rightarrow\left(ky-1\right)^2+4\left(k-1\right)=\left(ky-2\right)^2+\left(2ky-3\right)+4\left(k-1\right)>\left(ky-2\right)^2\)

\(\Rightarrow\left(ky-2\right)^2< \left(ky-1\right)^2+4\left(k-1\right)\le\left(ky-1\right)^2\)

\(\Rightarrow\left(ky-1\right)^2+4\left(k-1\right)=\left(ky-1\right)^2\)

\(\Rightarrow k=1\Rightarrow\dfrac{x^2+x+1}{xy+1}=1\)

\(\Rightarrow x^2+x=xy\Rightarrow y=x+1\)

\(\Rightarrow y-x=1\)