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Question 1

Question 2

Решите задачу:

$1)\;a)\;\frac{\sqrt{-125}}{\sqrt{-5}}\cdot\sqrt[3]8=\sqrt{\frac{-125}{-5}}\cdot\sqrt[3]{2^3}=\sqrt{25}\cdot2=5\cdot2=10$
$b)\;\frac{(\sqrt[4]{24}+\sqrt[4]6)^2}{4\sqrt3+3\sqrt6}=\frac{\sqrt[4]{24}\cdot\sqrt[4]{24}+\sqrt[4]{6}\cdot\sqrt[4]{24}+\sqrt[4]{6}\cdot\sqrt[4]{24}+\sqrt[4]{6}\cdot\sqrt[4]{6}}{4\sqrt3+3\sqrt6}=\frac{\sqrt[4]{24\cdot24}+2\sqrt[4]{6\cdot24}+\sqrt[4]{6\cdot6}}{4\sqrt3+3\sqrt6}=\frac{\sqrt{24}+2\sqrt[4]{144}+\sqrt{6}}{4\sqrt3+3\sqrt6}=\\=\frac{\sqrt{4\cdot6}+2\sqrt{12}+\sqrt6}{4\sqrt3+3\sqrt6}=\frac{2\sqrt6+4\sqrt3+\sqrt6}{4\sqrt3+3\sqrt6}=\frac{4\sqrt3+3\sqrt6}{4\sqrt3+3\sqrt6}=1$
$2.\;a)\;\frac{c+c^\frac12\cdot d^\frac12+d}{c^\frac32-d^\frac32}=\frac{c+c^\frac12\cdot d^\frac12+d}{(c^\frac12-d^\frac12)(c+c^\frac12\cdot d^\frac12+d)}=\frac1{c^\frac12-d^\frac12}$
$b)\;\frac{a-b}{a^\frac12-b^\frac12}-\frac{a^\frac32-b^\frac32}{a-b}=\frac{a-b}{a^\frac12-b^\frac12}-\frac{a^\frac32-b^\frac32}{(a^\frac12-b^\frac12)(a^\frac12+b^\frac12)}=\frac{(a^\frac12+b^\frac12)(a-b)-(a^\frac32-b^\frac32)}{{(a^\frac12-b^\frac12)(a^\frac12+b^\frac12)}}=$
$=\frac{a^\frac32-a^\frac12b+ab^\frac12-b^\frac32-a^\frac32+b^\frac32}{{(a^\frac12-b^\frac12)(a^\frac12+b^\frac12)}}=\frac{ab^\frac12-a^\frac12b}{{(a^\frac12-b^\frac12)(a^\frac12+b^\frac12)}}=\frac{a^\frac12b^\frac12(a^\frac12-b^\frac12)}{{(a^\frac12-b^\frac12)(a^\frac12+b^\frac12)}}=\frac{a^\frac12b^\frac12}{(a^\frac12+b^\frac12)}$