(x khác 0,-1)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2011}\Leftrightarrow-\frac{1}{x+1}=\frac{1}{2011}\Leftrightarrow x+1=-2011\Leftrightarrow x=-2012.\)

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

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

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

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

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

\(\Rightarrow2011.\left(-1\right)=\left(x+1\right).1\)

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

\(\Rightarrow x=-2011-1\)

\(\Rightarrow x=-2012\)

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

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

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

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

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

\(\Rightarrow x=-2011-1=-2012\)

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

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

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

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

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

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

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

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

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

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

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

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

                                          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 \(x=-2012\)

hnuji9on ui bm, 76tfv45tj,

Ta có \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xyz}\left(x+y+z\right)=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{xyz}=4\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)(vì \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>0\))

Mặt khác, ta có : \(\frac{1}{x+y+z}=2\) . 

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)

\(\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

=> x+y = 0 hoặc y + z = 0 hoặc z + x = 0

Từ đó suy ra P = 0 (lí do vì x,y,z là các số mũ lẻ)

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