Question

Чем сможете помогите. Заранее спасибо

Answer 1

$2)\; \; \frac{(x+2)^2}{x^2+5}-\frac{1+3x}{x^2+5}+\frac{2-x}{x^2+5}=\frac{x^2+4x+4-1-3x+2-x}{x^2+5}=\frac{x^2+6}{x^2+5}\\\\3)\; \; \frac{1}{x+1}-\frac{x-2}{3x^2+3x}=\frac{1}{x+1}-\frac{x-2}{3x(x+1)}=\frac{3x-x+2}{3x(x+1)}=\frac{2x+2}{3x(x+1)}=\\=\frac{2(x+1)}{3x(x+1)}=\frac{2}{3x}\\\\\frac{x-2}{x+x^2}+\frac{x}{x^2-1}=\frac{x-2}{x(x+1)}+\frac{x}{(x-1)(x+1)}=\frac{(x-2)(x-1)+x^2}{x(x+1)(x-1)}=\\=\frac{2x^2-3x+2}{x(x+1)(x-1)}\\\\\frac{2x^2}{x+1}-2x=\frac{2x^2-2x(x+1)}{x+1}=\frac{2x^2-2x^2-2x}{x+1}=-\frac{2x}{x+1}$

$5a)\; \; \frac{3x-3}{(x+1)(x-2)}=\frac{A}{x+1}+\frac{B}{x-2}=\frac{A(x-2)+B(x+1)}{(x+1)(x-2)}\\\\3x-3=A(x-2)+B(x+1)\\\\x=2:\; \; 3\cdot 2-3=0+B\cdot (2+1)\; ,\; \; 3=3B\; ,\; \; B=1\\\\x=-1:\; \; 3(-1)-3=A(-1-2)+0\; ,\; \; -6=-3A\; ,\; \; A=2\\\\\frac{3x-3}{(x+1)(x-2)}=\frac{2}{x+1}+\frac{1}{x-2}$

$5b)\; \; \frac{5x-4}{x^2-x-2}=\frac{5x-4}{(x+1)(x-2)}=\frac{A}{x+1}+\frac{B}{x-2}\\\\5x-4=A(x-2)+B(x+1)\\\\x=2:\; \; 5\cdot 2-4=3B\; ,\; B=2\\\\x=-1:\; \; -5-4=-3A\; ,\; \; A=3\\\\\frac{5x-4}{x^2-x-2}=\frac{3}{x+1}+\frac{2}{x-2}$