Ułamki i pierwiastki.

[jm]

\(\displaystyle{ \frac{2\sqrt{25}-\sqrt{27}}{\sqrt{27}-\sqrt{3}}-\frac{-5-13\sqrt{5}}{\sqrt{12}-\sqrt{27}}}\)

[Szemek]

\(\displaystyle{ \frac{10-3\sqrt{3}}{3\sqrt{3}-\sqrt{3}}-\frac{-5-13\sqrt{5}}{2\sqrt{3}-3\sqrt{3}}=}\)\(\displaystyle{ \frac{10-3\sqrt{3}}{2\sqrt{3}}-\frac{-5-13\sqrt{5}}{-\sqrt{3}}}\)
\(\displaystyle{ =\frac{10-3\sqrt{3}}{2\sqrt{3}}-\frac{-(5+13\sqrt{5})}{-\sqrt{3}}=}\)\(\displaystyle{ \frac{10-3\sqrt{3}}{2\sqrt{3}}-\frac{2(5+13\sqrt{5})}{2\sqrt{3}}=}\)
\(\displaystyle{ \frac{10-3\sqrt{3}}{2\sqrt{3}}-\frac{10+26\sqrt{5}}{2\sqrt{3}}=}\)\(\displaystyle{ \frac{-3\sqrt{3}-26\sqrt{5}}{2\sqrt{3}}=}\)\(\displaystyle{ \frac{-9-26\sqrt{15}}{6}}\)

[jm]

Zeby to bylo takie prose. Wynik musi wyjsc \(\displaystyle{ -\frac{29}{2}}\)

[Dargi]

\(\displaystyle{ \frac{2\sqrt{25}-\sqrt{27}}{\sqrt{27}-\sqrt{3}}-\frac{-5-13\sqrt{5}}{\sqrt{12}-\sqrt{27}}=
\frac{10-3\sqrt{3}}{3\sqrt{3}-\sqrt{3}}-\frac{-5-13\sqrt{5}}{2\sqrt{3}-3\sqrt{3}}=
\frac{10-3\sqrt{3}}{2\sqrt{3}}-\frac{-5-13\sqrt{5}}{-\sqrt{3}}=
\frac{10-3\sqrt{3}-10-26\sqrt{5}}{2\sqrt{3}}=\frac{-3\sqrt{3}-26\sqrt{5}}{2\sqrt{3}}=
\frac{(-3\sqrt{3}-26\sqrt{5})\sqrt{3}}{6}=\frac{(-9-26\sqrt{15}}{6}}\)

[MatizMac]

no mi to samo wychodzi, co Szemekowi Sprawdz moze jeszcze raz w odpowiedziach albo podaj nazwe dzialu, do ktorego odnosi sie to zadanie

[Szemek]

zamiast \(\displaystyle{ 13\sqrt{5}}\) powinno być \(\displaystyle{ 13\sqrt{3}}\)
i wtedy
\(\displaystyle{ \frac{-9-26\cdot 3}{6}=\frac{-87}{6}=-\frac{29}{2}}\)