Question

x^2+xy=12xy+y^2=4

x^2+xy=12
xy+y^2=4
x^2+y^2=26
xy=5
xy^2+xy+x=26
xy=6
另：解方程组时可以两式相除吗?（分母不为0）

Answer 1

x^2+xy=12
x(x+y)=12
xy+y^2=4
y(x+y)=4
x/y=3
x=3y
12y²=12
y=1 x=3
y=-1 x=-3
(2)x^2+y^2=26
xy=5
(x+y)²=x^2+y^2+2xy=36
x+y=±6
(x-y)²=x^2+y^2-2xy=16
x-y=±4
1)x+y=6 x-y=4
x=5 y=1
2)x+y=6 x-y=-4
x=1 y=5
3)x+y=-6 x-y=4
x=-1 y=-5
4)x+y=-6 x-y=-4
x=-5 y=-1
(3)xy^2+xy+x=26
x(y²+y+1)=26
xy=6
(y²+y+1)/y=13/3
3(y²+y+1)=13y
3y²-10y+3=0
1 -3
3 -1
(y-3)(3y-1)=0
y=3 x=2
y=1/3 x=18