Računanje s potencami s celimi eksponenti

$5^{2x+1}+2\cdot 5^{2x}-3\cdot 5^{2x-1}=5^{2x-1}(5^2+2\cdot 5^1-3)=$

$=5^{2x-1}\cdot 32=32\cdot 5^{2x-1}$

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

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

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

$\displaystyle \frac{1-3x^{-1}-10x^{-2}}{1-2x^{-1}-8x^{-2}}=\frac{1-\frac{3}{x}-\frac{10}{x^2}}{1-\frac{2}{x}-\frac{8}{x^2}}=\frac{\frac{x^2-3x-10}{x^2}}{\frac{x^2-2x-8}{x^2}}=$

$\displaystyle =\frac{\frac{(x-5)(x+2)}{x^2}}{\frac{(x-4)(x+2)}{x^2}}=\frac{(x-5)(x+2)x^2}{(x-4)(x+2)x^2}=\frac{x-5}{x-4}$

Nalogo lahko rešimo tudi tako, da števec in imenovalec ulomka pomnožimo z $x^2$.
$\displaystyle \frac{1-3x^{-1}-10x^{-2}}{1-2x^{-1}-8x^{-2}}=\frac{(1-3x^{-1}-10x^{-2})\cdot x^2}{(1-2x^{-1}-8x^{-2})\cdot x^2}=$
$\displaystyle =\frac{x^2-3x-10}{x^2-2x-8}=\frac{{(x-5)(x+2)}}{{(x-4)(x+2)}}=\frac{x-5}{x-4}$