\(\Rightarrow\frac{x}{2010}+\frac{x+1}{2011}+\frac{x+2}{2012}+\frac{x+3}{2013}+\frac{x+4}{2014}-5=0\)

\(\left(\frac{x}{2010}-1\right)+\left(\frac{x+1}{2011}-1\right)+\left(\frac{x+2}{2012}-1\right)\)\(+\left(\frac{x+3}{2013}-1\right)+\left(\frac{x+4}{2014}-1\right)=0\)

\(\frac{x-2010}{2010}+\frac{x-2010}{2011}+\frac{x-2010}{2012}+\frac{x-2010}{2013}+\frac{x-2010}{2014}=0\)

\(\left(x-2010\right).\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)=0\)

mà \(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\ne0\Rightarrow x+2010=0\Rightarrow x=-2010\)

Vậy x=-2010

\(x^2+y^2=1\Leftrightarrow\frac{^4}{a}+\frac{y^4}{b}=\frac{x^2+y^2}{a+b}\)

Theo tính chất tỉ lệ thức

\(\frac{x^2+y^2}{a+b}=\frac{x^2}{a}=\frac{y^2}{b}\left(a;b\ne0\right)\)

\(\frac{x^{2012}}{a^{1006}}+\frac{y^{2012}}{b^{1006}}=\left(\frac{x^2}{a}\right)^{1006}+\left(\frac{y^2}{b}\right)^{1006}=2.\left(\frac{x^2+y^2}{a+b}\right)^{2006}=\frac{2}{\left(a+b\right)^{2006}}\left(đpcm\right)\)

\(\frac{1}{x+1}+\frac{2}{x^2.\left(x-1\right)-\left(x-1\right)}+\frac{3}{\left(x-1\right).\left(x+1\right)}=0\)\(đk:x\ne\pm1\)

\(\frac{\left(x-1\right)^2}{\left(x+1\right).\left(x-1\right)^2}+\frac{2}{\left(x-1\right)^2.\left(x+1\right)}+\frac{3x-3}{\left(x-1\right)^2.\left(x+1\right)}=0\)

\(\Rightarrow\left(x-1\right)^2+2+3x-3=0\Rightarrow x.\left(x+1\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=-1\left(KTM\right)\end{cases}}\)

Vậy x=0

\(x^2-5x-9\)

\(=x^2-2.x.2,5+\left(2,5\right)^2-15,25\)

\(=\left(x-2,5\right)^2-\left(\sqrt{15,25}\right)^2\)

\(=\left(x-2,5-\sqrt{15,25}\right)\left(x-2,5+\sqrt{15,25}\right)\)

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

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

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