Question

Объясните, как делать номер 8 и 12.

Answer 1

$\boxed {(a^{n})^{k}=a^{nk}}\\\\8)\; \; (-2a^3b)^2=4\, a^6b^2\\\\(3xy^2)^4=81\, x^4y^8\\\\(\frac{1}{3}x^5y)^3=\frac{1}{27}\, x^{15}y^3\\\\(\frac{\sqrt2}{4}ab^5)^4=\frac{4}{256}\, a^4b^{20}=\frac{1}{64}\, a^4b^{20}$

$\boxed {a^{n}\cdot a^{k}=a^{n+k}\; \; ,\; \; a^{n}:a^{k}=a^{n-k}}\\\\12)\; \; 4x^5y^2:\, \boxed {A}=2x^3y\; \; \to \; \; \; \boxed {A}=\frac{4x^5y^2}{2x^3y}=2x^2y\\\\\boxed {A}:3ab^2=3ab\; \; \to \; \; \; \boxed {A}=3ab^2\cdot 3ab=9a^2b^3\\\\x^3y^8:\boxed {A}=3x\; \; \to \; \; \; \boxed {A}=\frac{x^3y^8}{3x}=\frac{x^2y^8}{3}=\frac{1}{3}x^2y^8\\\\\boxed {A}:a^2b^3=6a^3b^2\; \; \to \; \; \; \boxed {A}=a^2b^3\cdot 6a^3b^2=6a^5b^5$

Answer 2

Объяснение:

надеюсь понятно написала