Question

-6/x^2-1+3/x-1 при x=1999
x^2/x^2-2x+1-x+1/x-1
в)a^3+b^3/a^2-ab+b^2-a^3-b^3/a^2+ab+b^2 при a=14 15/16, b=0,05

Answer 1

а) $\mathtt{\frac{3}{x-1}-\frac{6}{x^2-1}=\frac{3(x+1)-6}{x^2-1}=\frac{3x-3}{x^2-1}=\frac{3}{x+1};~\frac{3}{1+1999}=0,0015}$

$\mathtt{\frac{x^2}{x^2-2x+1}-\frac{x+1}{x-1}=\frac{x^2-(x+1)(x-1)}{(x-1)^2}=\frac{1}{(x-1)^2}=}$

$\mathtt{\frac{a^3+b^3}{a^2-ab+b^2}-\frac{a^3-b^3}{a^2+ab+b^2}=\frac{(a+b)(a^2-ab+b^2)}{a^2-ab+b^2}-\frac{(a-b)(a^2+ab+b^2)}{a^2+ab+b^2}=2b=0,1}$