Question

помогите решить 45баллов

Answer 1

$1)\; \; -\dfrac{x^3}{x}=-x^2\\\\\dfrac{x\, (y+5)(y-1)}{x^2\, (y-1)}=\dfrac{y+5}{x}\\\\\\2)\; \; \dfrac{2a^2-8b^2}{a^2-4ab+4b^2}=\dfrac{2\, (a-2b)(a+2b)}{(a-2b)^2}=\dfrac{2\, (a+2b)}{a-2b}\\\\\\3)\; \; a=2\; ,\; b=1\\\\\dfrac{3a^2-6ab+3b^2}{4\, (a-b)(a+b)}=\dfrac{3\, (a-b)^2}{4\, (a-b)(a+b)}=\dfrac{3\, (a-b)}{4\, (a+b)}=\dfrac{3\cdot 1}{4\cdot 3}=\dfrac{1}{4}$

$4)\; \; \dfrac{28\, a^3\, b^{11}\, c^{23}}{56\, a^4\, b^{31}\, c^{12}}=\dfrac{c^{11}}{2\, a\, b^{20}}\\\\\\5)\; \; \dfrac{32\, a^2\, b^3\, c+16\, a^2\, b\, c-24\, a^3\, b^2\, c}{8a\, (4\, a\, b^2+2\, a-3\, a^2\, b)}=\dfrac{8a^2bc\, (4b^2+2-3ab)}{8a\cdot a(4b^2+2-3ab)}=\dfrac{b\, c}{1}=bc\\\\\\6)\; \; \dfrac{(a^{2n}-b^{2n})(a^{2n}-(ab)^{n}+b^{2n})}{a^{3n}+b^{3n}}=\dfrac{(a^{n}-b^{n})(a^{n}+b^{n})(a^{2n}-(ab)^{n}+b^{2n})}{(a^{n}+b^{n})(a^{2n}-(ab)^{n}+b^{2n})}=\\\\\\=a^{n}-b^{n}$