Question

Помогите с 1,4,6. Заранее спасибо

Answer 1

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

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

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