Question

Помогите сделать 1 2 5 и 7

Answer 1

Ответ на пятый задание :)

Answer 2

$1.\;\left(\frac x{xy-y^2}-\frac y{x^2-xy}\right):\frac{x^2-y^2}{8xy}=\left(\frac x{y(x-y)}-\frac y{x(x-y)}\right)\cdot\frac{8xy}{x^2-y^2}=\\\\\frac{x^2-y^2}{xy(x-y)}\cdot\frac{8xy}{x^2-y^2}=\frac8{x-y}$

$2.\;\frac{2^8\cdot7^9}{14^{10}}:\frac{26^5}{13^6\cdot8^4}=\frac{2^8\cdot7^9}{2^{10}\cdot7^{10}}\cdot\frac{13^6\cdot2^4\cdot4^4}{13^5\cdot2^5}=\frac1{2^2\cdot7^1}\cdot\frac{13^1\cdot4^4}{2^1}=\frac{13\cdot2^8}{2^3\cdot7}=\\\\=\frac{13\cdot2^5}{7}=\frac{13\cdot32}{7}=\frac{416}7=59\frac37$

$5.\;\frac{\sqrt[3]{250}}{{\sqrt[3]{2}}}=\sqrt[3]{\frac{250}2}=\sqrt[3]{125}=5\\\\\sqrt{75\cdot48}=\sqrt{25\cdot3\cdot48}=\sqrt{25\cdot144}=\sqrt{25}\cdot\sqrt{144}=5\cdot12=60$

$7.\;\frac{14}{1+5\sqrt2}=\frac{14\cdot\left(1-5\sqrt2\right)}{\left(1+5\sqrt2\right)\left(1-5\sqrt2\right)}=\frac{14\cdot\left(1-5\sqrt2\right)}{1-50}=\frac{14\cdot(5\sqrt2-1)}{49}=\frac{2\cdot\left(5\sqrt2-1\right)}{7}=\frac{10\sqrt2-2}7$