Odpowiedź:
SZALKI
Szczegółowe wyjaśnienie:
Skorzystam z twierdzenia:
[tex]\sqrt{a}\cdot\sqrt{a}=a,\ a\geq0[/tex]
[tex]\dfrac{3}{\sqrt{15}}=\dfrac{3}{\sqrt{15}}\cdot\dfrac{\sqrt{15}}{\sqrt{15}}=\dfrac{3\!\!\!\!\diagup^1\sqrt{15}}{15\!\!\!\!\!\diagup_5}=\dfrac{\sqrt{15}}{5}\qquad\boxed{\bold{S}}\\\\\dfrac{15}{\sqrt3}=\dfrac{15}{\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}=\dfrac{15\!\!\!\!\!\diagup^5\sqrt3}{3\!\!\!\!\diagup_1}=5\sqrt3\qquad\boxed{\bold{Z}}\\\\\dfrac{1}{\sqrt{15}}=\dfrac{1}{\sqrt{15}}\cdot\dfrac{\sqrt{15}}{\sqrt{15}}=\dfrac{\sqrt{15}}{15}\qquad\boxed{\bold{A}}[/tex]
[tex]\dfrac{15}{\sqrt{15}}=\dfrac{15}{\sqrt{15}}\cdot\dfrac{\sqrt{15}}{\sqrt{15}}=\dfrac{15\!\!\!\!\!\diagup\sqrt{15}}{15\!\!\!\!\!\diagup}=\sqrt{15}\qquad\boxed{\bold{L}}\\\\\dfrac{5}{\sqrt{15}}=\dfrac{5}{\sqrt{15}}\cdot\dfrac{\sqrt{15}}{\sqrt{15}}=\dfrac{5\!\!\!\!\diagup^1\sqrt{15}}{15\!\!\!\!\!\diagup_3}=\dfrac{\sqrt{15}}{3}\qquad\boxed{\bold{K}}\\\\\dfrac{15}{\sqrt{5}}=\dfrac{15}{\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}=\dfrac{15\!\!\!\!\!\diagup^3\sqrt5}{5\!\!\!\!\diagup_1}=3\sqrt5\qquad\boxed{\bold{I}}[/tex]
