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\(A=\dfrac{2012\left(2012+1\right)+2012+2}{2012-2-2012\cdot2015}\)
\(=\dfrac{2012^2+2012\cdot2+2}{2012-2-2012\left(2012+3\right)}\)
\(=\dfrac{2012^2+4026}{-2012^2-4026}=-1\)

a/ \(\frac{2.2012.2014-2}{2011+2012.\left(2014-1\right)}=\frac{2.\left(2012.2014-1\right)}{2011+2012.2014-2012}\)
\(=\frac{2.\left(2012.2014-1\right)}{2012.2014-1}=2\)
b/ \(\frac{2012.2013+2014}{2010-2012.\left(2013+2\right)}=\frac{2012.2013+2014}{2010-2012.2013-4024}\)
\(=\frac{2012.2013+2014}{-\left(2012.2013+2014\right)}=-1\)
c/ \(\frac{6.11111.87564-3.11111}{2.11111\left(87564-4\right)+7.11111}=\frac{6.87564-3}{2.87564-8+7}\)
\(=\frac{3\left(2.87564-1\right)}{2.87564-1}=3\)

\(a,\frac{3.\left(x-y\right)}{y-x}=\frac{-3.\left(y-x\right)}{y-z}=-3\)
\(b,\frac{x^2-x}{1-x}=\frac{x.\left(x-1\right)}{1-x}=\frac{-x.\left(1-x\right)}{1-x}=-x\)
\(\frac{3\left(x-y\right)}{y-x}=\frac{3\left(x-y\right)}{-1\left(x-y\right)}=-3\)
\(\frac{x^2-x}{1-x}=\frac{x\left(x-1\right)}{-1\left(x-1\right)}=-x\)

\(\frac{x^2+2x+1}{5x^3+5x^2}=\frac{\left(x+1\right)^2}{5x^2\left(x+1\right)}=\frac{x+1}{5x^2};\)
b, \(\frac{2x^2+2x}{x+1}=\frac{2x\left(x+1\right)}{x+1}=2x\)

\(a,\frac{4x^3}{10x^2y}=\frac{2x}{5y}\)
\(b,\frac{10xy^5\left(2x-3y\right)}{12xy\left(2x-3y\right)}=\frac{5y^4}{6}\)
Hok Tốt~~
\(\frac{4x^3}{10x^2y}=\frac{2x}{5y}\)
\(\frac{10xy^5\left(2x-3y\right)}{12xy\left(2x-3y\right)}=\frac{5y^4}{4}\)
Tham khảo nhé~

\(\frac{x^2-5x+6}{x^2-2x}=\frac{x^2-2x-3x+6}{x.\left(x-2\right)}=\frac{x.\left(x-2\right)-3.\left(x-2\right)}{x.\left(x-2\right)}\)
\(=\frac{\left(x-3\right).\left(x-2\right)}{x.\left(x-2\right)}=\frac{x-3}{x}\)

a)\(\dfrac{12x^3y^2}{18xy^5}\)=\(\dfrac{2x^2}{3y^3}\)
b)\(\dfrac{15x.\left(x+5\right)^2}{20x^2.\left(x+5\right)}\)=\(\dfrac{3.5x\left(x+5\right)}{4x.5x.\left(x+5\right)}\)=\(\dfrac{3\left(x+5\right)}{4x}\)

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

a)\(\dfrac{3x^2-12x+12}{x^4-8x}=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}=\dfrac{3\left(x-2\right)^2}{x\left(x^3-2^3\right)}=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
\(A=\frac{2012.2013+2014}{2010-2012.2015}\)
\(=\frac{2012.2013+2014}{2010-2012.\left(2013+2\right)}\)
\(=\frac{2012.2013+2014}{2010-2012.2013-4024}\)
\(=\frac{2012.2013+2014}{-2012.2013-2014}=-1\)