Question

Обчисліть значення виразу $\frac{3\sqrt{2}-5 }{\sqrt{2} -1} +\frac{\sqrt{24}-\sqrt{300} }{\sqrt{3} }$

Answer 1

Решение:

$\frac{3\sqrt{2}-5}{\sqrt{2}-1}+\frac{\sqrt{24}-\sqrt{300}}{\sqrt{3}}=\frac{3\sqrt{2}-5}{\sqrt{2}-1}*\frac{\sqrt{2}+1}{\sqrt{2}+1}+\frac{\sqrt{4*6}-\sqrt{100*3}}{\sqrt{3}}=\frac{(3\sqrt{2}-5)(\sqrt{2}+1)}{2-1}+\frac{2\sqrt{6}-10\sqrt{3}}{\sqrt{3}}=\frac{(3\sqrt{2}-5)(\sqrt{2}+1)}{1}+\frac{2\sqrt{6}-10\sqrt{3}}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}=6+3\sqrt{2}-5\sqrt{2}-5+\frac{(2\sqrt{6}-10\sqrt{3})*\sqrt{3}}{3}=1-2\sqrt{2}+\frac{2\sqrt{18}-10*3}{3}=1-2\sqrt{2}+\frac{2\sqrt{9*2}-30}{3}=$

$=1-2\sqrt{2}+\frac{6\sqrt{2}-30}{3}=1-2\sqrt{2}+\frac{3(2\sqrt{2}-10)}{3}=1-2\sqrt{2}+2\sqrt{2}-10=-9$

Ответ: -9

$DK954$