Question

помогите упростить выражение,номер 3​

Answer 1

$\displaystyle\bf\\1)\\\\\dfrac{b^{\frac{1}{3} } }{b-1} +\dfrac{b}{b^{\frac{4}{3} } -b^{\frac{2}{3} } }=\frac{b^{\frac{1}{3} } }{b-1} +\frac{b}{b^{\frac{2}{3} }(b^{\frac{2}{3} } -1) } =\\\\\\=\frac{b^{\frac{1}{3} } }{(b^{\frac{1}{3} } -1)(b^{\frac{2}{3} } +b^{\frac{1}{3} } +1)} +\frac{b}{b^{\frac{2}{3} } (b^{\frac{1}{3} } -1)(b^{\frac{1}{3} } +1)}} =$

$\displaystyle\bf\\=\frac{b^{\frac{1}{3} } \cdot(b^{\frac{1}{3} }+1)+b^{\frac{1}{3} } \cdot(b^{\frac{2}{3} } +b^{\frac{1}{3} } +1) }{ (b-1)(b^{\frac{2}{3} } -1)} =\frac{b^{\frac{1}{3} } (b^{\frac{2}{3} } +2b^{\frac{1}{3} } +2)}{(b-1)(b^{\frac{2}{3} } -1)} \\\\\\2)\\\\\frac{b^{\frac{1}{3} } \Big(b^{\frac{2}{3} } +2b^{\frac{1}{3} } +2\Big)}{(b-1)(b^{\frac{2}{3} } -1)} \cdot\Big(b^{\frac{2}{3} } -1\Big)\cdot\frac{b-1}{b^{\frac{1}{3} } } =b^{\frac{2}{3} } +2b^{\frac{1}{3} } +2$