Question

помогите пожалуйста срочно​

Answer 1

Ответ:

$\frac{3(a-b)(a+b)}{6(a+b)(a-b)}=\frac{3}{6}=\frac{1}{2}\\\\\\\frac{3(n^2+n+1)}{(n-1)(n^2+n+1)}=\frac{3}{n-1}\\\\\\\frac{12a^3b^5(p-q)^2}{36a^2b(q-p)^2}=\frac{ab^4(p-q)(p-q)}{3(q-p)(q-p)} =\frac{ab^4(p^2-2pq+q^2)}{3(q^2-2qp+p^2)}=\frac{ab^4(p^2-2pq+q^2)}{3(p^2-2pq+q^2)} =\frac{ab^4}{3}$

$\frac{49xy(c-d)^2}{7x^2(d-c)^2} =\frac{7y(c^2-2cd+d^2)}{x(c^2-2cd+d^2)}=\frac{7y}{x} \\\\\\\frac{c(z-15)^3}{8c(z-15)^4} =\frac{(z-15)(z-15)(z-15)}{8(z-15)(z-15)(z-15)(z-15)}=\frac{1}{8(z-15)} (=\frac{1}{8z-120})\\\\\\\frac{3a(b-2)}{6(b-2)}=\frac{a(b-2)}{2(b-2)(b-2)}=\frac{a}{2(b-2)}$