\(3,\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)^2-\frac{x^2+y^2}{\left(x-y\right)^2}\)

\(=\frac{2}{xy}:\left[\left(\frac{1}{x}\right)^2-2.\frac{1}{x}.\frac{1}{y}+\left(\frac{1}{y}\right)^2\right]-\frac{x^2+y^2}{\left(x-y\right)^2}\)

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

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

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

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

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

\(\frac{2011^3+11^3}{2011^3+2000^3}\)

\(=\frac{\left(2011+11\right)\left(2011^2-2011.11+11^2\right)}{\left(2011+2000\right)\left(2011^2-2011.2000+2000^2\right)}\)

\(=\frac{\left(2011+11\right)\left[2011^2-11\left(2011-11\right)\right]}{\left(2011+2000\right)\left[2011^2-2000\left(2011-2000\right)\right]}\)

\(=\frac{\left(2011+11\right)\left(2011^2-11.2000\right)}{\left(2011+2000\right)\left(2011^2-2000.11\right)}\)

\(=\frac{2011+11}{2011+2000}\left(2011^2-11.2000\ne0\right)\)

                                          đpcm

\(A=\left(\frac{a+1}{ab+1}+\frac{ab+a}{ab-1}-1\right):\left(\frac{a+1}{ab+1}-\frac{ab+a}{ab-1}+1\right)\)

\(A=\left[\frac{\left(a+1\right)\left(ab-1\right)+\left(ab+a\right)\left(ab+1\right)-\left(ab+1\right)\left(ab-1\right)}{\left(ab+1\right)\left(ab-1\right)}\right]:\left[\frac{\left(a+1\right)\left(ab-1\right)-\left(ab+a\right)\left(ab+1\right)+\left(ab+1\right)\left(ab-1\right)}{\left(ab+1\right)\left(ab-1\right)}\right]\)\(A=\left[\frac{a^2b-a+ab-1+a^2b^2+ab+a^2b+a-a^2b^2+1}{\left(ab+1\right)\left(ab-1\right)}\right]:\left[\frac{a^2b-a+ab-1-a^2b^2-ab-a^2b-a+a^2b^2-1}{\left(ab+1\right)\left(ab-1\right)}\right]\)\(A=\left[\frac{2a^2b+2ab}{\left(ab+1\right)\left(ab-1\right)}\right]:\left[\frac{2a^2b-2a}{\left(ab+1\right)\left(ab-1\right)}\right]\)

\(A=\left[\frac{2ab\left(a+1\right)}{\left(ab+1\right)\left(ab-1\right)}\right]:\left[\frac{2a\left(ab-1\right)}{\left(ab+1\right)\left(ab-1\right)}\right]\)

\(A=\left[\frac{2ab\left(a+1\right)}{\left(ab+1\right)\left(ab-1\right)}\right]:\left[\frac{2a}{\left(ab+1\right)}\right]\left(ab-1\ne0\right)\)

\(A=\frac{b\left(a+1\right)}{ab-1}\left(ab+1\ne0;2a\ne0\right)\)

Ta có:\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}+\frac{1}{2011}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2011}\)

\(\Leftrightarrow-\frac{1}{x+1}=\frac{1}{2011}\)\(\Leftrightarrow-x-1=2011\)

\(\Leftrightarrow x=-2012\)

\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}+\frac{1}{2011}\)

=> \(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}-\frac{-1}{2011}\)

=> \(\frac{1}{x+1}=\frac{-1}{2011}=\frac{1}{-2011}\)

=> x + 1 = -2011

=> x = -2011 - 1

=> x = -2012

Vậy x = -2012

Ta có: \(\frac{1}{x\left(x+1\right)}=\frac{1}{x}+\frac{1}{2011}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2011}\)

\(\Leftrightarrow\frac{1}{x+1}=-\frac{1}{2011}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{-2011}\)

\(\Leftrightarrow x+1=-2011\)

\(\Leftrightarrow x=-2012\)

Vậy \(x=-2012\)

ĐKXĐ: \(x\ne0,x\ne-1\)

Ngoài việc quy đồng có thể giải như sau:

Ta thấy: \(\frac{1}{x\left(x+1\right)}=\frac{\left(x+1\right)-x}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)

Nên từ đề bài => \(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2011}\)

=>\(-\frac{1}{x+1}=\frac{1}{2011}\)=> \(-\left(x+1\right)=2011\)=>\(-x-1=2011\)=>\(x=-2012\)( thỏa mãn ĐKXĐ)

Kết luận.

Ta có \(A=[\frac{2}{\left(x+1\right)^3}\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}\left(\frac{1}{x^2}+1\right)]:\frac{x-1}{x^3}\)

\(\Leftrightarrow A=\left[\frac{2}{\left(x+1\right)^3}.\frac{x+1}{x}+\frac{1}{\left(x+1\right)^2}.\frac{x^2+1}{x^2}\right].\frac{x^3}{x-1}\)

\(\Leftrightarrow A=\left[\frac{2x+x^2+1}{x^2\left(x+1\right)^2}\right].\frac{x^3}{x+1}=\frac{x}{x+1}\)

Để \(A=\frac{x}{x+1}< 1\Leftrightarrow\frac{1}{x+1}>0\Leftrightarrow x>-1\)

Để \(A=1-\frac{1}{x+1}\text{ nguyên thì }\frac{1}{x+1}\text{ nguyên hay }x\in\left\{-2,0\right\} \)

\(\frac{1}{x\left(x+1\right)}=\frac{\left(x+1\right)-x}{x\left(x+1\right)}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)

=>\(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2011}\)

=>\(\frac{1}{x}-\frac{1}{x+1}-\frac{1}{x}=\frac{1}{2011}\)

=>\(\frac{1}{x}-\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2011}\)

=>\(0-\frac{1}{x+1}=\frac{1}{2011}\)

=>\(-\frac{1}{x+1}=\frac{1}{2011}\)

=>-x+1=2011

=>-x=2011-1

=>-x=2010

=>x=-2010

Vậy x=-2010

\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}+\frac{1}{2011}\)

<=>\(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2011}\)

<=>\(-\frac{1}{x+1}=\frac{1}{2011}\)

<=>-x-1=2011

<=>x=-2012

Đáp số: \(x=-2012\)

<=>\(\frac{1}{x}-\frac{1}{x+1}\)=\(\frac{1}{x+}+\frac{1}{2011}\)

<=>\(\frac{1}{x}+\left(\frac{-1}{x+1}\right)=\frac{1}{x}+\frac{1}{2011}\)

<=>\(-\frac{1}{x}=\frac{1}{2011}\)

<=>2011(-1)=x.1

<=>-2011=x

<=>x=-2011

vậy x=-2011