Question

Помогите мне пж с Кр дам 135 баллов

Answer 1

Объяснение:

1.а) $\frac{14a^3b^5}{21a^4b}=\frac{2b^4}{3a}$

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

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

2.а) $\frac{2x}{x-a}-\frac{2a}{x+a}=$$\frac{2x(x+a)-2a(x-a)}{(x-a)(x+a)}=\frac{2x^2+2ax-2ax+2a^2}{x^2-a^2}=\frac{2^2+2a^2}{x^2-a^2}$

б) $\frac{2-ab}{2a+ab}+\frac{2b}{2+b}=\frac{2-ab}{a(2+b)}+\frac{2b}{2+b}=\frac{2-ab+2ab}{a(2+b)}=\frac{2+ab}{2a+ab}$

в) $c-\frac{c^2}{c+1}=\frac{c(c+1)-c^2}{c+1}=\frac{c^2+c-c^2}{c+1}=\frac{c}{c+1}$

3. $\frac{7}{x^2-y^2}-\frac{5}{xy-x^2}-\frac{12}{x^2+xy}=\frac{7}{(x-y)(x+y)}-\frac{5}{x(y-x)}-\frac{12}{x(x+y)}=\frac{7}{(x-y)(x+y)}-\frac{5}{x(-(x-y))}-\frac{12}{x(x+y)}=\frac{7}{(x-y)(x+y)}+\frac{5}{x(x-y)}-\frac{12}{x(x+y)}=\frac{7x+5(x+y)-12(x-y)}{x(x-y)(x+y)}=\frac{7x+5x+5y-12x+12y}{x(x^2-y^2)}=\frac{0+17y}{x^3-xy^2}=\frac{17y}{x^3-xy^2}$

4. $\frac{ax-ay+3x-3y}{a^2-9}=\frac{a(x-y)+3(x-y)}{(a-3)(a+3)}=\frac{(x-y)(a+3)}{(a-3)(a+3)}=\frac{x-y}{a-3}=\frac{5,8-3,4}{3,1-3}=\frac{2,4}{0,1}=24$