Question

ДОПОМОЖІТЬ, БУДЬ ЛАСКА!
1) 8+a³/16-a⁴ : a²-2a+4/a²+4 =
2) (5x²/8y³)³ * (-16y⁵/25x³)² =
3) a²-4p²/9a²-p² : a²+4ap+4p²/9a²-6ap+p² * 3a+p/a-2p =​

Answer 1

$\displaystyle\bf\\1)\\\\\frac{8+a^{3} }{16-a^{4} } :\frac{a^{2} -2a+4}{a^{2} +4} =\frac{2^{3} +a^{3} }{4^{2} -(a^{2} )^{2} } :\frac{a^{2} -2a+4}{a^{2} +4} =\\\\\\=\frac{(a+2)(a^{2} -2a+4)}{(4-a^{2})(4+a^{2} ) } \cdot\frac{a^{2} +4}{a^{2} -2a+4} =\frac{a+2}{4-a^{2} } =\\\\\\=\frac{a+2}{(2-a)(2+a)} =\frac{1}{2-a} \\\\\\2)$

$\displaystyle\bf\\\Bigg(\frac{5x^{2} }{8y^{3} } \Bigg)^{3} \cdot\Bigg(-\frac{16y^{5} }{25x^{3} } \Bigg)^{2} =\frac{125x^{6} }{512y^{9} } \cdot\frac{256y^{10} }{625x^{6} } =\frac{y}{2\cdot 5} =\frac{y}{10} \\\\\\3)\\\\\frac{a^{2} -4p^{2} }{9a^{2}-p^{2} } :\frac{a^{2} +4ap+4p^{2} }{9a^{2} -6ap+p^{2} } \cdot\frac{3a+p}{a-2p} =\\\\\\=\frac{(a -2p)(a+2p) }{(3a-p^)(3a+p) } :\frac{(a+2p)^{2} }{(3a -p)^{2} } \cdot\frac{3a+p}{a-2p} =$

$\displaystyle\bf\\=\frac{(a -2p)(a+2p) }{(3a-p^)(3a+p) } \cdot\frac{(3a-p)^{2} }{(a+ 2p)^{2} } \cdot\frac{3a+p}{a-2p} =\frac{3a-p}{a+2p}$