Usuń niewymierność z mianownika.
a) 3/√15
b) 10/√5
c) 9/2√3
d) 8/3√2 ​



Odpowiedź :

Obliczenia do a) i b)

[tex]\frac{3}{\sqrt{15}}=\frac{3}{\sqrt{15}}\cdot\frac{\sqrt{15}}{\sqrt{15}}=\frac{3\sqrt{15}}{\sqrt{15^2}}=\frac{3\sqrt{15}}{15}=\frac{\sqrt{15}}{5}\\\\\frac{10}{\sqrt5}=\frac{10}{\sqrt5}\cdot\frac{\sqrt5}{\sqrt5}=\frac{10\sqrt5}{\sqrt{5^2}}=\frac{10\sqrt5}{5}=2\sqrt5[/tex]

Obliczenia do c) i d)

[tex]\frac{9}{2\sqrt3}=\frac{9}{2\sqrt3}\cdot\frac{\sqrt3}{\sqrt3}=\frac{9\sqrt3}{2\sqrt{3^2}}=\frac{9\sqrt3}{2\cdot3}=\frac{3\sqrt3}{2}\\\\\frac{8}{3\sqrt2}=\frac{8}{3\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}=\frac{8\sqrt2}{3\sqrt{2^2}}=\frac{8\sqrt2}{3\cdot2}=\frac{4\sqrt2}{3}[/tex]