Question

1.
(x2-1)/(1-x3)-x2/(x2+x+1)
2.
(x4-y4)/(x3+yx2-xy2-y3)-2y
3.
a/(a2-3a)+(a-8)/a+(7a2-3a-72)/(a3-9a)
помогите пожалуйста

Answer 1

1. 
$frac{x^2-1}{1-x3}-frac{x^2}{x^2+x+1}=frac{x^2-1}{(1-x)(1+x+x^2)}-frac{x^2}{x^2+x+1}=frac{x^2-1-x^2+x^3}{(1-x)(x^2+x+1)}=\=frac{x^3-1}{1-x^3}=-1$

2.
$frac{x^4-y^4}{x^3+yx^2-xy^2-y^3}-2y=frac{(x^2-y^2)(x^2+y^2)}{x^2(x+y)-y^2(x+y)}-2y=frac{(x^2-y^2)(x^2+y^2)}{(x+y)(x^2-y^2)}-2y=\=frac{x^2+y^2}{x+y}-2y=frac{x^2+y^2-2xy-2y^2}{x+y}=frac{x^2-2xy-y^2}{x+y}$

3.
$frac{a}{a^2-3a}+frac{a-8}{a}+frac{7a^2-3a-72}{a^3-9a}=frac{a}{a(a-3)}+frac{a-8}{a}+frac{7a^2-3a-72}{a(a-3)(a+3)}=\= frac{a(a+3)+(a-8)(a^2-9)+7a^2-3a-72}{a(a-3)(a+3)}=\= frac{a^2+3a+a^3-9a-8a^2+72+7a^2-3a-72}{a(a-3)(a+3)}=frac{a^3-9a}{a(a-3)(a+3)}=frac{a(a-3)(a+3)}{a(a-3)(a+3)}=1$