Question

Помогите с 1,4,6. Заранее спасибо

Answer 1

$1); ; frac{2a}{a+2}+frac{2-a}{a+2}=frac{2a+2-a}{a+2}=frac{a-2}{a+2}\\frac{2x+2}{2x-1}+frac{x-1}{1-2x}=frac{2x+2}{2x-1}-frac{x-1}{2x-1}=frac{2x+2-x+1}{2x-1}=frac{x+3}{2x-1}\\frac{2-y}{(3-y)^2}-frac{5-2y}{(y-3)^2}=[(3-y)^2=(y-3)^2]=frac{2-y-5+2y}{(y-3)^2}=frac{y-3}{(y-3)^2}=frac{1}{y-3}$

$4); ; frac{1}{(x-1)x}+frac{1}{x(x+1)}+frac{1}{(x+1)(x+2)}+frac{1}{(x+2)(x+3)}=\\=frac{x+1+x-1}{(x-1), x, (x+1)}+frac{x+3+x+1}{(x+1)(x+2)(x+3)}=frac{2x}{(x-1), x, (x+1)}+frac{2x+4}{(x+1)(x+2)(x+3)}=\\= frac{2}{(x-1)(x+1)}+frac{2(x+2)}{(x+1)(x+2)(x+3)}=frac{2}{(x-1)(x+1)}+frac{2}{(x+1)(x+3)}=\\=frac{2(x+3)+2(x-1)}{(x-1)(x+1)(x+3)}=frac{4(x+1)}{(x-1)(x+1)(x+3)}=frac{4}{(x-1)(x+3)}$

$6); ; frac{3n^2-2n-3}{n}=frac{3n^2}{n}-frac{2n}{n}-frac{3}{n}=3n-2-frac{3}{n}\\frac{3n^2+4n-2}{n+1}=frac{(3n^2+3n)+(n+1)-3}{n+1}=frac{3n(n+1)+(n+1)-3}{n+1}=\\=frac{3n(n+1)}{n+1}+frac{n+1}{n+1}-frac{3}{n+1}=3n+1-frac{3}{n+1}$

Answer 2

Большое спасибо!