Question

Алгебра . Пожалуйста!!!ПжПж​

Answer 1

Ответ:

скачай приложение "photomath",оно тебе решит.

Answer 2

Объяснение:

$a) \ \frac{\sqrt{2}+\sqrt{8}-\sqrt{32} }{-\sqrt{2} } =\frac{\sqrt{2}+2\sqrt{2}-4\sqrt{2} }{-\sqrt{2} }= \frac{\sqrt{2}*(1+2-4) }{-\sqrt{2} } =\frac{-\sqrt{2} }{-\sqrt{2} }=1.\\ b)\ \frac{\sqrt{3}-\sqrt{27}+\sqrt{48} }{\sqrt{3} } =\frac{\sqrt{3}-3\sqrt{3}+4\sqrt{3} }{\sqrt{3} } =\frac{\sqrt{3}*(1-3+4) }{\sqrt{3} }=2.\\ c)\ \frac{10-\sqrt{125} }{\sqrt{5*(2-\sqrt{5})^2 } } =\frac{10-5\sqrt{5} }{|2-\sqrt{5}|*\sqrt{5}}=\frac{-5(\sqrt{5}-2) }{(\sqrt{5}-2)*\sqrt{5} } =-\sqrt{5}.$

$d)\ \frac{(\sqrt{2}+1)^3+(\sqrt{2}-1)^3 }{\sqrt{8} } =\frac{(\sqrt{2}+1+\sqrt{2}-1)*(\sqrt{2}+1)^2-(\sqrt{2}+1)*(\sqrt{2}-1)+(\sqrt{2}-1)^2 }{2\sqrt{2} }=\\ =\frac{2\sqrt{2} *(2+2\sqrt{2} +1-2+1+2-2\sqrt{2} +1)}{2\sqrt{2} } =5.$