Question

8 КЛАСС, ДАЮ 35 БАЛЛОВ!!!

Answer 1

Ответ:

$1)\ \ \displaystyle \frac{a^3b}{15c}\cdot \Big(-\frac{3c}{a^2b^2}\Big)=-\frac{a}{5\, b}\\\\\\18y^3\cdot \frac{4x^2}{9y^5}=\frac{2\cdot 4x^2}{y^2}=\frac{8x^2}{y^2}\\\\\\\frac{a^{3n}\, b^{n+4}}{c^{n+3}}\cdot \frac{c^{2n+3}}{a^{2n}\, b^{n+1}}=\frac{a^{n}\, b^3\, c^{n}}{1}=a^{n}\, b^3\, c^{n}$

$2)\ \ \displaystyle \Big(-\frac{3a}{2b^2}\Big)^4= \frac{81a^4}{16b^8}\\\\\\\Big(-\frac{5\, a^3\, b^4}{3\, c^5\, d^7}\Big)^3=-\frac{125\, a^9\, b^{12}}{27\, c^{15}\, d^{21}}\\\\\\\\3)\ \ \frac{18m^3n^4}{25\, p^6\, q^{10}}:\Big(-\frac{4\, m^2\, n^9}{75\, p^5\, q^{12}}\Big)=\frac{18\, m^3n^4\cdot 75\, p^5\, q^{12}}{25\, p^6\, q^{10}\cdot 4\, m^2\, n^9}=\frac{9\, m\cdot 3q^2}{p\cdot 2n^5}=\frac{27\, m\, q^2}{2\, p\, n^5}$

$\dfrac{48\, x^4y^3}{49\, z^9}:(16x^7y^8)=\dfrac{48\, x^4\, y^3}{49\, z^9\cdot 16\, x^7\, y^8}=\dfrac{3}{49\, z^9\, x^3\, y^5}$