Question

представьте в виде дроби выражения. Нужно б в и г.​

Answer 1

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

$\displaystyle\bf\\2)\\\\\frac{1}{c^{2}-4c } -\frac{1}{c^{2}-8c+16 } =\frac{1}{c\cdot(c-4)} -\frac{1}{(c-4)^{2} } =\\\\\\=\frac{c-4-c}{c\cdot(c-4)^{2} }=-\frac{4}{c(c-4)^{2} } \\\\\\3)\\\\\frac{m^{3}-n^{3} }{m+n}-m^{2} + n^{2} =\frac{m^{3}-n^{3}-(m^{2} -n^{2} )\cdot(m+n) }{m+n} =\\\\\\=\frac{m^{3}-n^{3}-m^{3}-m^{2} n+mn^{2} +n^{3} }{m+n}=\\\\\\=\frac{-m^{2}n+mn^{2} }{m+n} =\frac{mn(n-m)) }{m+n}$