ПОВТОРЮЄМО АЛГЕБРУ ЗА 8 КЛАС: 17

17. Виконайте дію:

1) $\frac{{2a}^2-2a}{{4a}^2-b^2}$ + $\frac{b(a-1)}{b^2-{4a}^2}$ = $\frac{{2a}^2-2a}{{4a}^2-b^2}$ – $\frac{b(a-1)}{{4a}^2-b^2}$ = $\frac{2a(a-1)-b(a-1)}{{4a}^2-b^2}$ = $\frac{(a-1)(2a-b)}{(2a-b)(2a+b)}$ = $\frac{a-1}{2a+b}$;

2) $\frac{x+3}{\mathrm{xy}-x^2}$ + $\frac{y+3}{\mathrm{xy}-y^2}$ = $\frac{x+3}{x(y-x)}$ + $\frac{y+3}{y(x-y)}$ = $\frac{{x+3}^{\backslash y}}{x(y-x)}$ – $\frac{{y+3}^{\backslash x}}{y(y-x)}$ = $\frac{\mathrm{xy}+3y-\mathrm{xy}-3x}{\mathrm{xy}(y-x)}$ = $\frac{-3(x-y)}{\mathrm{xy}(y-x)}$ = $\frac{-3(x-y)}{-\mathrm{xy}(x-y)}$ = $\frac{3}{\mathrm{xy}}$;

3) $\frac{x^2+x-3}{x^2-9}$ – $\frac{3}{2x-6}$ – 1 = $\frac{x^2+x-3^{\backslash 2}}{(x-3)(x+3)}$ – $\frac{3^{\backslash x+3}}{2(x-3)}$ $1^{\backslash 2(x^2-9)}$ = $\frac{{2x}^2+2x-6-3x-9-{2x}^2+18}{2(x-3)(x+3)}$ = $\frac{-x+3}{2(x-3)(x+3)}$ = $\frac{-(x-3)}{2(x-3)(x+3)}$ = –$\frac{1}{2(x+3)}$;

4) $\frac{\mathrm{ax}-2x}{3y+6}$ : $\frac{{4-a}^2}{y^2-4y+4}$ = $\frac{x(a-2)}{3(y+2)}$ • $\frac{{(y+2)}^2}{(2+a)(2+a)}$ = –$\frac{x(a-2)(y+2)}{3(a-2)(a+2)}$ = –$\frac{x(y+2)}{3(a+2)}$.