Question

№4. Срочно!.........

Answer 1

a) $3\sqrt{2\frac{1}{3} } -\sqrt{84} -4\sqrt{5\frac{1}{4} } =\\= 3\sqrt{\frac{7}{3} } -2\sqrt{21} -4\sqrt{\frac{21}{4} } =\\= 3*\frac{\sqrt{7} }{\sqrt{3} } -2\sqrt{21} -4*\frac{\sqrt{21} }{2} =\\=\frac{3\sqrt{7} }{\sqrt{3} } -2\sqrt{21} -2\sqrt{21} = \sqrt{21} -2\sqrt{21} -2\sqrt{21} =\\ =-3\sqrt{21}$

б) $(27b^-^6^)^-^\frac{2}{3}=(27*\frac{1}{b^6} )^-^\frac{2}{3} = (\frac{27}{b^6} )^-^\frac{2}{3}=(\frac{b^6}{27} )^\frac{2}{3}=\sqrt[3]{(\frac{b^6}{27})^2 } =\sqrt[3]{\frac{b^6}{27}^2 } =\\=(\frac{b^2}{3} )^2 = \frac{b^4}{9}$

в) $\sqrt[5]{\frac{a^2\sqrt[4]{a^-^3} }{a^-^\frac{1}{4}}}= \sqrt[5]{\frac{a^\frac{5}{4}}{a^-^\frac{1}{4}}} = \sqrt[5]{a^\frac{3}{2}} =(a^\frac{3}{2} )^\frac{1}{5} =a^\frac{3}{10}=\sqrt[10]{a^3}$

г) $\frac{3a}{a^2-9} -\frac{3}{a^2-9} :(\frac{a+2}{3a-3} -\frac{1}{a-1} )= \\\frac{3a}{a^2-9} -\frac{3}{a^2-9} :(\frac{a+2}{3*(a-1)} -\frac{1}{a-1})=\frac{3a}{a^2-9} -\frac{3}{a^2-9} : \frac{a+2-3}{3*(a-1)} =\\ \frac{3a}{a^2-9}- \frac{3a}{a^2-9}:\frac{a-1}{3*(a-1)} = \frac{3a}{a^2-9}- \frac{3a}{a^2-9}:\frac{1}{3} = \frac{3a}{a^2-9}- \frac{3a}{a^2-9} * 3 = \\\frac{3a}{a^2-9}- \frac{9}{a^2-9} = \frac{3a-9}{a^2-9} = \frac{3*(a-3)}{(a-3)*(a+3)} =\\ \frac{3}{a+3}$