Question

надо решить номера 2, 4, 6, 8, 10, 14; ​

Answer 1

$2)\ \ \dfrac{a-2b}{x+c}\cdot \dfrac{3x+3c}{5a-10b}=\dfrac{a-2b}{x+c}\cdot \dfrac{3\, (x+c)}{5\, (a-2b)}=\dfrac{3}{5}\\\\\\4)\ \ \dfrac{ab-2b^2}{xy+y^2}\cdot \dfrac{5y+5x}{2b^2-ab}=\dfrac{ab-2b^2}{y\, (x+y)}\cdot \dfrac{5\, (x+y)}{-(ab-2b^2)}=-\dfrac{5}{y}$

$6)\ \ \dfrac{8a+3b}{3c}:\dfrac{3b+8a}{3n}=\dfrac{8a+3b}{3c}\cdot \dfrac{3n}{8a+3b}=\dfrac{n}{c}\\\\\\8)\ \ \dfrac{x^2-c^2}{a-b}:\dfrac{x+c}{a^2-b^2}=\dfrac{(x-c)(x+c)}{a-b}\cdot \dfrac{(a-b)(a+b)}{x+c}=(x-c)(a+b)$

$10)\ \ \dfrac{5a+3x}{2x}\cdot 3x=\dfrac{3\cdot (5a+3x)}{2}\\\\\\11)\ \ \dfrac{8y}{y^2-a^2}\cdot (y+a)=\dfrac{8y}{(y-a)(y+a)}\cdot (y+a)=\dfrac{8y}{y-a}$

Answer 2

Жауабы осы болады........