Алгебра, вопрос задал sazonovaa695 , 7 лет назад

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

Ответы на вопрос

Ответил tamarabernukho
0

 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)
Новые вопросы