Question

Нужна ваша помощь. Алгебра 7 класс. Нужны ответы на 1, 2, 4, 6.

Answer 1

Ответ:

1) $\frac{12-2x}{x-2}$+$\frac{10-x}{2-x}$ = $\frac{12-2x}{x-2}$+$\frac{10-x}{-(x-2)}$ = $\frac{12-2x-(10-x)}{x-2}$ = $\frac{12-2x-10+x}{x-2}$ = $\frac{2-x}{x-2}$ = $\frac{-(x-2)}{x-2}$ = -1.

2) $\frac{12p^3-1}{3p^2}$-$\frac{1-3p^3}{3p^2}$ = $\frac{12p^3-1-(1-3p^3)}{3p^2}$ = $\frac{12p^3-1-1+3p^3}{3p^2}$ = $\frac{15p^3-2}{3p^2}$.

4) $\frac{5x+1}{5x-20}$+$\frac{x+17}{20-5x}$ = $\frac{5x+1}{5x-20}$-$\frac{x+17}{5x-20}$ = $\frac{5x+1=(x+17)}{5x-20}$ = $\frac{5x+1-x-17}{5(x-4)}$ =  $\frac{4x-16}{5(x-4)}$ = $\frac{4(x-4)}{5(x-4)}$ = $\frac{4}{5}$.

6) $\frac{64-2ab}{(a-8)^2}$+$\frac{2ab-a^2}{(8-a)^2}$ = $\frac{64-2ab}{(a-8)^2}$+$\frac{2ab-a^2}{(a-8)^2}$ = $\frac{64-2ab+2ab-a^2}{(a-8)^2}$ = $\frac{-(a-8)*(8+a)}{(a-8)^2}$ = $\frac{-(8+a)}{a-8}$ = $-\frac{8+a}{a-8}$.