Решите систему уравнений 1) {(x+y)/(x-y)+x/y=-5/6 {(x^2+xy)/(xy-y^2)=1/6 2) {корень из (x+3y)/(y+5)+2=3 корня из (y+5)/(x+3y) {xy+2x=13-4y

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Ответ:

fariza34

08.10.2020 08:49

$\left \{ {{ \frac{x+y}{x-y} + \frac{x}{y} =- \frac{5}{6} } \atop { \frac{x^2+xy}{xy-y^2} = \frac{1}{6} }} \right. \\ \\ ODZ:x \neq y;y \neq 0 \\ \\ \left \{ {{ \frac{x+y}{x-y} + \frac{x}{y} =- \frac{5}{6} } \atop { \frac{x(x+y)}{y(x-y)} = \frac{1}{6} }} \right. \\ \\ a= \frac{x+y}{x-y} ;b= \frac{x}{y} \\ \\ \left \{ {{a+b=- \frac{5}{6} } \atop {ab= \frac{1}{6} }} \right. \\ \\ \left \{ {{b=-a- \frac{5}{6} } \atop {a(-a- \frac{5}{6}) = \frac{1}{6} }} \right. \\ \\ 6a^2+5a+1=0 \\ \\$
$D=1 \\ \\ a_{1} =- \frac{1}{2 } \\ b_{1} =- \frac{1}{3} \\ \\ \left \{ {{ \frac{x+y}{x-y} =- \frac{1}{2} } \atop { \frac{x}{y} =- \frac{1}{3} }} \right. \\ \\ y=-3x;x \neq 0 \\ \\ a_{2} =- \frac{1}{3 } \\ b_{2} =- \frac{1}{2} \\ \\ \left \{ {{ \frac{x+y}{x-y} =- \frac{1}{3} } \atop { \frac{x}{y} =- \frac{1}{2} }} \right. \\ \\ y=-2x;x \neq 0 \\ \\ OTVET: \\ \\ y=-3x;x \neq 0 \\ \\ y=-2x;x \neq 0 \\ \\ \\$

$\left \{ {{ \sqrt{ \frac{x+3y}{y+5} } +2=3 \sqrt{ \frac{y+5}{x+3y} } } \atop {xy+2x=13-4y}} \right. \\ \\ y \neq -5;x \neq -3y \\ \\ t= \sqrt{ \frac{x+3y}{y+5} } ;t\ \textgreater \ 0 \\ \\ t+2= \frac{3}{t} \\ \\ t^2+2t-3=0 \\ \\ D=16 \\ \\ t_{1} =(-2-4)/2=-3\ \textless \ 0; \\ \\ t_{2} =(-2+4)/2=1\ \textgreater \ 0 \\ \\ t=1 \\ \\$
$\left \{ {{ \sqrt{ \frac{x+3y}{y+5} } =1} \atop {xy+2x=13-4y}} \right. \\ \\ \left \{ {{x+3y=y+5} \atop {xy+2x=13-4y}} \right. \\ \\ \left \{ {{x=-2y+5} \atop {-4y+10-2y^2+5y=13-4y}} \right. \\ \\ \left \{ {{x=-2y+5} \atop {2y^2-5y+3=0}} \right. \\ \\ D=1 \\ \\ y_{1} =(5-1)/4=1 \\ x_{1} =-2+5=3 \\ \\ y_{2} =(5+1)/4=1.5 \\ x_{2} =2 \\ \\ OTVET:(3;1)(2;1.5)$