решите систему уравнений
А) (x/x+3) + (x+3/x)=17/4
x^2-4xy+4y^2=0
Б)(y+1/y+2)+(y+2/y+1)=25/12
x^2-2xy+y^2=0

1{x²-4xy+4y²=0⇒(x-2y)²=0⇒x-2y=0⇒x=2y{x/(x+3)+(x+3)/x=17/4⇒2y/(2y+3)+(2y+3)/2y=17/4(2y+3)/2y=a1/a+a=17/44a²-17a+4=0D=289-64=225a1=(17-15)/8=1/4⇒(2y+3)/2y=1/48y+12=2y6y=-12y1=-2⇒x1=-4a2=(17+15)/8=4⇒(2y+3)/2y=42y+3=8y6y=3y2=0,5⇒x2=1(-4;-2);(1;0,5)2{x²-2xy+y²=0⇒(x-y)²=0⇒x-y=0⇒x=y{(y+1)/(y+2)+(y+2)/(y+1)=25/12(y+2)/(y+1)=aa+1/a=25/1212a²-25a+12=0D=625-576=49a1=(25-7)/24=3/4⇒(y+2)/(y+1)=3/44y+8=3y+3y1=-5⇒x1=-5a2=(25+7)/24=4/3⇒(y+2)/(y+1)=4/33y+6=4y+4y2=2⇒x2=2(-5;-5);(2;2)