Question

Помогите, максимально баллов дам​

Answer 1

Ответ:

Объяснение:

$\displaystyle\\\left \{ {{\dfrac{xy}{x+y}=\bigg{-6} }\atop {\Big\\\dfrac{xz}{x+z} =\dfrac{-10}{3} }} \atop {\dfrac{yz}{y+z} =\dfrac{15}{8} }} \right. <=>\left \{ {{\dfrac{x+y}{xy}=\bigg{-\dfrac{1}{6} } }\atop {\Big\\\dfrac{x+z}{xz} =\dfrac{3}{-10} }} \atop {\dfrac{y+z}{yz} =\dfrac{8}{15} }} \right.<=>\left \{ {{\dfrac{1}{y} +\dfrac{1}{x} =-\dfrac{1}{6} }\atop {\Big\\\dfrac{1}{z} +\dfrac{1}{x} =\dfrac{-10}{3} }} \atop {\dfrac{1}{z}+\dfrac{1}{y} =\dfrac{8}{15} }} \right.$

$\displaystyle\left \{ {{\dfrac{1}{x} =-\dfrac{1}{6}- \dfrac{1}{y} }\atop {\Big\\\dfrac{8}{15}- \dfrac{1}{y}-\dfrac{1}{6}- \dfrac{1}{y} =\dfrac{-10}{3} }} \atop {\dfrac{1}{z} =\dfrac{8}{15}- \dfrac{1}{y}}} \right.$

решим второе уравнение

$\displaystyle\\\frac{2}{y}=\frac{8}{15}-\frac{1}{6} +\frac{3}{10} =\frac{32-10+18}{60} =\frac{40}{60}=\frac{2}{3};\ \ \ \ y=3$

$\displaystyle\\\frac{1}{x} =-\frac{1}{6} -\frac{1}{3} =-\frac{3}{6}= -\frac{1}{2};\ \ \ \ \ \ x=-2\\\\\\\frac{1}{z}=\frac{8}{15}-\frac{1}{3}=\frac{3}{15}=\frac{1}{5} ;\ \ \ \ \ \ \ \ \ \ z=5$

Ответ: $x=-2;y=3;z=5$