Помогите решить систему уравнений

Ответ

Проверено экспертом

Автор - Universalka

$left { {{(x+0,2)^{2} +(y+0,3)^{2}=1} atop {x+y=0,9}} right.\\left { {{(x+0,2)^{2} +(y+0,3)^{2}=1} atop {(x+0,2)+(y+0,3)=1,4}} right.\\x+0,2=m\y+0,3=n\\left { {{m^{2}+n^{2}=1} atop {m+n=1,4}} right.\\left { {{(m+n)^{2}-2mn=1} atop {m+n=1,4}} right.$

$left { {{1,4^{2} -2mn=1} atop {m+n=1,4}} right.\\left { {{2mn=0,96} atop {m+n=1,4}} right.\\left { {{m*n=0,48} atop {m+n=1,4}} right.\\1)left { {{m=0,6} atop {n=0,8}} right.\\left { {{x+0,2=0,6} atop {y+0,3=0,8}} right.\\left { {{x_{1}=0,4 } atop {y_{1}=0,5 }} right.$

$2)left { {{m=0,8} atop {n=0,6}} right.\\left { {{x+0,2=0,8} atop {y+0,3=0,6}} right.\\left { {{x_{2}=0,6 } atop {y_{2}=0,3 }} right.\\Otvet:(0,4;0,5),(0,6;0,3)$

$left { {{x-y=1} atop {x^{3}-y^{3}=7}} right.\\:left { {{(x-y)(x^{2}+xy+y^{2})=7} atop {x-y=1}} right.\---------\x^{2}+xy+y^{2}=7\\left { {{x=1+y} atop {x^{2}+xy+y^{2}=7}} right.\\left { {{x=y+1} atop {(y+1)^{2}+y(y+1)+y^{2}=7}} right.\\left { {{x=y+1} atop {y^{2}+2y+1+y^{2}+y+y^{2}=7}} right.\\left { {{3y^{2}+3y-6=0} atop {x=y+1}} right.\\left { {{y^{2}+y-2=0} atop {x=y+1}} right.$

$1)left { {{y_{1}=-2 } atop {x=-2+1}} right.\\left { {{y_{1} =-2} atop {x_{1} =-1}} right.\\2)left { {{y_{2}=1 } atop {x_{2} =1+1}} right.\\left { {{y_{2}=1 } atop {x_{2}=2 }} right.\\Otvet:(-1;-2),(2;1)$

$left { {{x^{-1} +y^{-1}=5} atop {x^{-2}+y^{-2}=13}} right.\\x^{-1}=mRightarrow x^{-2}=m^{2}\y^{-1}=nRightarrow y^{-2}=n^{2}\\left { {{m+n=5} atop {m^{2} +n^{2}=13 }} right.\\-left { {{m^{2}+2mn+n^{2}=25} atop {m^{2}+n^{2}=13}} right.\---------\2mn=12\m*n=6\\left { {{m+n=5} atop {m*n=6}} right.\\1)left { {{m=2} atop {n=3}} right. \\left { {{x^{-1}=2 } atop {y^{-1}=3 }} right.\\left { {{x=frac{1}{2} } atop {y=frac{1}{3}}} right.$

$2)left { {{m=3} atop {n=2}} right.\\left { {{x^{-1}=3} atop {y^{-1} =2}} right.\\left { {{x_{2}=frac{1}{3}} atop {y_{2}=frac{1}{2}}} right.\\Otvet:(frac{1}{2};frac{1}{3}),(frac{1}{3};frac{1}{2})$