Question

Решите еее 30 баллов

Answer 1

Ответ:

$-\frac{10y^{3}}{9x^{2}};$ $-\frac{3p+q}{4pq};$ $\frac{2}{(a-7b)(2a+3b)};$ $\frac{c-d}{c+d};$

Объяснение:

$a) -\frac{14x^{3}*45y^{7}}{27y^{4}*21x^{5}}=-\frac{7x^{3}*2*5y^{3}*9y^{4}}{3*9y^{4}*7x^{3}*3x^{2}}=-\frac{2*5y^{3}}{3*3x^{2}}=-\frac{10y^{3}}{9x^{2}};$

$b) \frac{(3p-q)(3p+q)}{4pq(q-3p)}=-\frac{(q-3p)(3p+q)}{4pq(q-3p)}=-\frac{3p+q}{4pq};$

$c) \frac{2(2a-3b)(a-7b)}{(a-7b)(a-7b)(2a-3b)(2a+3b)}=\frac{2}{(a-7b)(2a+3b)};$

$d) \frac{(c-d)(c^{2}+cd+d^{2})(c+d)}{(c+d)(c+d)(c^{2}+cd+d^{2})}=\frac{c-d}{c+d};$