Question

Нужно очень срочно, дам 30 баллов

Answer 1

$1)\; \; \dfrac{1}{x^5}=x^{-5}\; \; ,\; \; \; \; \dfrac{1}{8}=8^{-1}\\\\\\2)\; \; 2^{-7}\cdot 2^6=\dfrac{1}{2^7}\cdot 2^6=\dfrac{2^6}{2^7}=2\\\\5:5^{-2}=5:\dfrac{1}{5^2}=5\cdot 5^2=5^3=125\\\\\Big((\dfrac{1}{4})^{-1}\Big)^{2}=4^2=16\\\\\\3)\; \; 2x^{-3}y^2(3x^{-2}y^{-4})=2\cdot \dfrac{1}{x^3}\cdot y^2\cdot 3\cdot \dfrac{1}{x^2}\cdot \dfrac{1}{y^4}=\dfrac{6\, y^2}{x^5\, y^4}=\dfrac{6}{x^5\, y^2}\\\\\\4)\; \; (\dfrac{1}{2}xy^{-3})^{-2}=\dfrac{1}{2^{-2}}\cdot x^{-2}\cdot y^{6}=\dfrac{4\, y^6}{x^2}$

$5)\; \; (x^{-2}-y^{-2})(x+y)^{-1}=\Big(\dfrac{1}{x^2}-\dfrac{1}{y^2}\Big)\cdot\dfrac{1}{x+y}=\Big(\dfrac{1}{x}-\dfrac{1}{y}\Big)\Big(\dfrac{1}{x}+\dfrac{1}{y}\Big)\cdot \dfrac{1}{x+y}=\\\\\\=\dfrac{y-x}{xy}\cdot \dfrac{x+y}{xy}\cdot \dfrac{1}{x+y}=\dfrac{y-x}{x^2y^2}$