Question

допоможіть будь-ласка дам 100 балів!!​

Answer 1

Пояснення:

Вправа 1.

4) $(a - b)^{\frac{4}{9} }$ = $\sqrt[9]{(a - b)^{4} }$

5) $p^{\frac{1}{4} } - b^{\frac{3}{4} }$ = $\sqrt[4]{p} - \sqrt[4]{b^{3} }$

Вправа 2.

2) $121^{\frac{1}{2} }$ = $\frac{1}{121^{\frac{1}{2} } }$ = $\frac{1}{(11^{2})^{\frac{1}{2} } }$ = $\frac{1}{11}$

3) $16^{\frac{1}{4} }$ = $(2^{4)^{\frac{1}{4} } }= 2$

4) $0,008^{\frac{2}{3} } = (\frac{1}{125})^{\frac{2}{3} }= \sqrt[3]{(\frac{1}{125})^{2} } = \sqrt[3]{\frac{1}{125} } ^{2} = (\frac{1}{5})^{2} = \frac{1}{25}$ або 0,04

5) $0,0016^{-\frac{1}{4} } = (\frac{1}{626})^{-\frac{1}{4} } = 625^{-\frac{1}{4} } = (5^{4})^{\frac{1}{4} } = 5$

6) $0,36^{0,5} = (\frac{9}{25} )^{\frac{1}{2} }= \sqrt{\frac{9}{25} } = \frac{3}{5}$ або 0,6

7) $(11\frac{1}{9} )^{-1,5} = (11\frac{1}{9})^{-\frac{3}{2} } = \frac{1}{\sqrt{(11\frac{1}{9})^{\frac{3}{2} } } } = \frac{1}{(11\frac{1}{9})^{3} } = \frac{1}{\frac{100}{9}*\sqrt{\frac{100}{9} } } = \frac{1}{\frac{100}{9}*\frac{10}{3} } = \frac{1}{\frac{1000}{27} } = \frac{27}{1000}$ або 0,027

8) $(4\frac{17}{27} )^{-\frac{2}{3} } = \frac{1}{(4\frac{17}{27})^{\frac{2}{3} } } = \frac{1}{\sqrt[3]{(4\frac{17}{27})^{2} } } = \frac{\sqrt[3]{4\frac{17}{27} } }{\frac{125}{27} } = \frac{\sqrt[3]{\frac{125}{27} } }{y\frac{125}{27} } = \frac{\frac{5}{3} }{\frac{125}{27} } = \frac{9}{25}$ або 0,36

Вправа 3.

1) $b^{\frac{1}{3} } * b^{\frac{1}{2} } = b^{\frac{5}{6} } = \sqrt[6]{b^{5} }$

2) $b^{-\frac{1}{10} }b^{\frac{1}{2} }= b^{\frac{2}{5} } = \sqrt[5]{b^{2} }$

3) $b^{-2}b^{-\frac{1}{8} } = b^{-2} * \frac{1}{b^{\frac{1}{8} } } = \frac{b^{-2} }{b^{\frac{1}{8} } } = \frac{1}{b^{\frac{17}{8} } } = \frac{1}{\sqrt[8]{b^{17} } } = \frac{1}{b^{2}\sqrt[8]{b} } = \frac{1}{b^{2}\sqrt[8]{b} }* \frac{\sqrt[8]{b^{7} } }{\sqrt[8]{b^{7} } }= \frac{1\sqrt[8]{b^{7} } }{b^{2}\sqrt[8]{b}\sqrt[8]{b^{7} } } = \frac{\sqrt[8]{b^{7} } }{b^{2}\sqrt[8]{b*b^{7} } } = \frac{\sqrt[8]{b^{7} } }{b^{2}\sqrt[8]{b^{8} } } = \frac{\sqrt[8]{b^{7} } }{b^{2}*b } =$$= \frac{\sqrt[8]{b^{7} } }{b^{3} }$

4) $b^{\frac{1}{4} } : b^{\frac{1}{8} } = b^{\frac{1}{8} } = \sqrt[8]{b}$

6) $b^{-\frac{3}{8} } : b^{-\frac{1}{24} } = \frac{\sqrt[3]{b^{2} } }{\sqrt[3]{b}^{3} } = \frac{\sqrt[3]{b^{2} } }{b}$

5) $b^{\frac{1}{5} } / b^{-\frac{1}{15} } = b^{\frac{4}{15} } = \sqrt[15]{b^{4} }$

7) $(b^{-\frac{1}{3}})^{3} = b^{-1} = \frac{1}{b}$

8) $(b^{\frac{3}{5} } )^{-\frac{10}{27} } = b^{\frac{2}{9} } = \sqrt[9]{b^{2} }$