Uzasadnij, że podana liczba jest całkowita.

${\left({\left(\frac{1}{2}\right)}^0-\sqrt{2}\cdot 0,25^{-1,25}\right)}^{-1}\cdot \frac{\sqrt{2}\left(3^{24}-2\cdot 3^{23}+7\cdot 3^{22}\right)}{\sqrt{16200}\cdot 3^{18}\cdot {\left(0,(14)+1,(41)\right)}^{-1}}$

Rozwiązanie

${\left({\left(\frac{1}{2}\right)}^0-\sqrt{2}\cdot 0,25^{-1,25}\right)}^{-1}={\left(1-\sqrt{2}\cdot {\left(\frac{1}{4}\right)}^{-1\frac{1}{4}}\right)}^{-1}={\left(1-\sqrt{2}\cdot 4^{\frac{5}{4}}\right)}^{-1}={\left(1-2^{\frac{1}{2}}\cdot {\left(2^2\right)}^{\frac{5}{4}}\right)}^{-1}=$
$={\left(1-2^{\frac{1}{2}}\cdot 2^{\frac{5}{2}}\right)}^{-1}={\left(1-2^{\frac{6}{2}}\right)}^{-1}={\left(1-2^3\right)}^{-1}={\left(1-8\right)}^{-1}={\left(-7\right)}^{-1}=-\frac{1}{7}$

$0,(14)=0,141414\dots =x$
$0,141414\dots =x|\cdot 100$
$14,141414\dots =100x|-x$
$14=99x|:99$
$x=\frac{14}{99}⟶0,(14)=\frac{14}{99}$

Analogicznie: $0,(41)=\frac{41}{99}$, więc $1,(41)=1\frac{41}{99}$