a) $2011.2013+2012.2014$

$=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)$

$=2012^2-1+2013^2-1$

$=2012^2+2013^2-2$

$\Rightarrow2011.2013+2012.2014=2012^2+2013^2-2$

b) $\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)$

$=\dfrac{1}{10}\left(9+1\right)\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)$

$=\dfrac{1}{10}\left(9^2-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)$

$=\dfrac{1}{10}\left(9^4-1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)$

$=\dfrac{1}{10}\left(9^8-1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)$

$=\dfrac{1}{10}\left(9^{16}-1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)$

$=\dfrac{1}{10}\left(9^{32}-1\right)\left(9^{32}+1\right)$

$=\dfrac{1}{10}\left(9^{64}-1\right)$

$=\dfrac{9^{64}-1}{10}$

Ta có: $9^{64}-1=\dfrac{10\left(9^{64}-1\right)}{10}$

Mà $\dfrac{10\left(9^{64}-1\right)}{10}>\dfrac{9^{64}-1}{10}$

$\Rightarrow\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)< 9^{64}-1$

c) Ta có:

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

Vì x>y>0, ta có:

$\dfrac{x-y}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}\left(2\right)$

Vì x>y>0 nên $\left(x+y\right)^2-xy< \left(x+y\right)^2\left(3\right)$

Từ (1), (2) và (3) suy ra:

$\dfrac{x-y}{x+y}< \dfrac{x^2-y^2}{x^2+xy+y^2}$