Помогите пожалуйста, решите первый вариант,очень нужно.

Question 1

Question 2

1.
a) $\frac{12}{x^2-4}+ \frac{3}{x+2}$ = $\frac{12}{(x-2)(x+2)} + \frac{3}{(x+2)}$ = $\frac{12+3(x-2)}{(x-2)(x+2)}$ = $\frac{12+3x-6}{(x-2)(x+2)}$ = $\frac{6+3x}{(x-2)(x+2)}$ = $\frac{3(2+x)}{(x-2)(x+2)}$ = $\frac{3}{x-2}$ = $\frac{3}{2002-2}$ = $\frac{3}{2000}$ = $0.0015$
б) $\frac{x^2}{x^2+2x+1} - \frac{x-1}{x+1}$ = $\frac{x^2}{(x+1)^2} - \frac{x-1}{x+1}$ = $\frac{x^2-(x+1)(x-1)}{(x+1)^2}$ = $\frac{x^2-(x^2-1)}{(x+1)^2}$ = $\frac{x^2-x^2+1}{(x+1)^2}$ = $\frac{1}{(x+1)^2}$ = $\frac{1}{(19+)^2}$ = $\frac{1}{20^2}$ = $\frac{1}{400}$ = $0.0025$
в) $\frac{a^3-b^3}{a^2+ab+b^2} + \frac{a^3+b^3}{z^2-ab+b^2}$ = $\frac{(a-b)(a^2+ab+b^2)}{a^2+ab+b^2}+ \frac{(a+b)(a^2-ab+b^2)}{a^2-ab+b^2}$ = $a-b+a+b$ = $2a$ = $2*0.05$ = $0.1$

2.
$\frac{2}{x-2} - \frac{12}{x^3-8} - \frac{x-2}{x^2+2x+4}$ = $\frac{2}{x-2} - \frac{12}{(x-2)(x^2+2x+4)} - \frac{x-2}{x^2+2x+4}$ = $\frac{2(x^2+2x+4)-12-(x-2)(x-2)}{(x-2)(x^2+2x+4)}$ = $\frac{2x^2+4x+8-12-(x-2)^2}{x^3-8}$ = $\frac{2x^2+4x+8-12-x^2+4x-4}{x^3-8}$ = $\frac{x^2+8x-8}{X^3-8}$

a)x=0
$\frac{x^2+8x-8}{x^3-8}$ = $\frac{0^2+8*0-8}{0^3-8}$ = $\frac{0+0-8}{0-8}$ = $\frac{-8}{-5}$ = $1$

б) $\frac{x^2+8x-8}{x^3-8}$ = $\frac{2^2+8*2-8}{2^3-8} [/tex=[tex] \frac{4+16-8}{8-8}$ = $\frac{12}{0}$

Question 3

Спасибо