Odpowiedź :
a)
[tex]\sqrt{5}(3-\sqrt{5})=3\sqrt{5}-\sqrt{5}*\sqrt{5}=\boxed{3\sqrt{5}-5}[/tex]
b)
[tex]\sqrt{2}(5\sqrt{2}-3)=\sqrt{2}*5\sqrt{2}-3\sqrt{2}=\boxed{10-3\sqrt{2}}[/tex]
c)
[tex]-\sqrt{3}(2\sqrt{3}+5)=-\sqrt{3}*2\sqrt{3}-5\sqrt{3}=\boxed{-6-5\sqrt{3}}[/tex]
d)
[tex]2\sqrt{7}(4+\sqrt{7})=2\sqrt{7}*4+2\sqrt{7}*\sqrt{7}=\boxed{8\sqrt{7}+14}[/tex]
e)
[tex]3\sqrt{5}(3-2\sqrt{5})=3\sqrt{5}*3-3\sqrt{5}*2\sqrt{5}=\boxed{9\sqrt{5}-30}[/tex]
f)
[tex]-4\sqrt{2}(3\sqrt{2}-1)=-4\sqrt{2}*3\sqrt{2}-4\sqrt{2}*(-1)=\boxed{-24+4\sqrt{2}}[/tex]
Odpowiedź:
[tex]a)\ \ \sqrt{5}(3-\sqrt{5})=3\sqrt{5}-\sqrt{5}\cdot\sqrt{5}=3\sqrt{5}-5\\\\b)\ \ \sqrt{2}(5\sqrt{2}-3)=\sqrt{2}\cdot5\sqrt{2}-\sqrt{2}\cdot3=2\cdot5-3\sqrt{2}=10-3\sqrt{2}\\\\c)\ \ -\sqrt{3}(2\sqrt{3}+5)=-\sqrt{3}\cdot2\sqrt{3}-\sqrt{3}\cdot5=-3\cdot2-5\sqrt{3}=-6-5\sqrt{3}\\\\d)\ \ 2\sqrt{7}(4+\sqrt{7})=2\sqrt{7}\cdot4+2\sqrt{7}\cdot\sqrt{7}=8\sqrt{7}+2\cdot7=8\sqrt{7}+14[/tex]
[tex]e)\ \ 3\sqrt{5}(3-2\sqrt{5})=3\sqrt{5}\cdot3-3\sqrt{5}\cdot2\sqrt{5}=9\sqrt{5}-6\cdot5=9\sqrt{5}-30\\\\f)\ \ -4\sqrt{2}(3\sqrt{2}-1)=-4\sqrt{2}\cdot3\sqrt{2}-4\sqrt{2}\cdot(-1)=-12\cdot2+4\sqrt{2}=-24+4\sqrt{2}\\\\\\Wynik\ \ mno\.zenia\ \ tego\ \ samego\ \ pierwiastka\ \ drugiego\ \ stopnia\ \ jest\\\\r\'owny\ \ liczbie\ \ podpierwiastkowej\\\\\sqrt{5}\cdot\sqrt{5}=5\\\\\sqrt{2}\cdot\sqrt{2}=2\\\\\sqrt{3}\cdot\sqrt{3}=3[/tex]