Question

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

Answer 1

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=a
1/a+a=17/4
4a²-17a+4=0
D=289-64=225
a1=(17-15)/8=1/4⇒(2y+3)/2y=1/4
8y+12=2y
6y=-12
y1=-2⇒x1=-4
a2=(17+15)/8=4⇒(2y+3)/2y=4
2y+3=8y
6y=3
y2=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)=a
a+1/a=25/12
12a²-25a+12=0
D=625-576=49
a1=(25-7)/24=3/4⇒(y+2)/(y+1)=3/4
4y+8=3y+3
y1=-5⇒x1=-5
a2=(25+7)/24=4/3⇒(y+2)/(y+1)=4/3
3y+6=4y+4
y2=2⇒x2=2
(-5;-5);(2;2)