解方程(x2+1)*(y2+1)=10、(x+y)*(xy-1)=3
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/25 01:44:26
解方程(x2+1)*(y2+1)=10、(x+y)*(xy-1)=3
(x²+1)(y²+1)=10
x²y²+x²+y²+1=10
x²y²-2xy+1+x²+y²+2xy=10
(xy-1)²+(x+y)²=10
[(xy-1)+(x+y)]²-2(x+y)(xy-1)=10
(x+y)(xy-1)=3代入
[(xy-1)+(x+y)]²=16
(xy-1)+(x+y)=4或(xy-1)+(x+y)=-4
x+y和xy-1是方程t²-4t+3=0或t²+4t+3=0的根.
t²-4t+3=0 (t-1)(t-3)=0 t=1或t=3
x+y=1 xy-1=3 xy=4 x,y是方程m²-m+4=0的根,方程无实根.
x+y=3 xy-1=1 xy=2 x,y是方程m²-3m+2=0的根.
(m-1)(m-2)=0 m=1或m=2
x=1 y=2或x=2 y=1
t²+4t+3=0 (t+1)(t+3)=0 t=-1或t=-3
x+y=-1 xy-1=-3 xy=-2 x,y是方程m²+m-2=0的根,
(m-2)(m+1)=0 m=2或m=-1
x=2 y=-1或x=-1 y=2
x+y=-3 xy-1=-1 xy=0 x,y是方程m²+3m=0的根.
m(m+3)=0 m=0或m=-3
x=0 y=-3或x=-3 y=0
综上,得:
x=1 y=2
x=2 y=1
x=2 y=-1
x=-1 y=2
x=0 y=-3
x=-3 y=0
x²y²+x²+y²+1=10
x²y²-2xy+1+x²+y²+2xy=10
(xy-1)²+(x+y)²=10
[(xy-1)+(x+y)]²-2(x+y)(xy-1)=10
(x+y)(xy-1)=3代入
[(xy-1)+(x+y)]²=16
(xy-1)+(x+y)=4或(xy-1)+(x+y)=-4
x+y和xy-1是方程t²-4t+3=0或t²+4t+3=0的根.
t²-4t+3=0 (t-1)(t-3)=0 t=1或t=3
x+y=1 xy-1=3 xy=4 x,y是方程m²-m+4=0的根,方程无实根.
x+y=3 xy-1=1 xy=2 x,y是方程m²-3m+2=0的根.
(m-1)(m-2)=0 m=1或m=2
x=1 y=2或x=2 y=1
t²+4t+3=0 (t+1)(t+3)=0 t=-1或t=-3
x+y=-1 xy-1=-3 xy=-2 x,y是方程m²+m-2=0的根,
(m-2)(m+1)=0 m=2或m=-1
x=2 y=-1或x=-1 y=2
x+y=-3 xy-1=-1 xy=0 x,y是方程m²+3m=0的根.
m(m+3)=0 m=0或m=-3
x=0 y=-3或x=-3 y=0
综上,得:
x=1 y=2
x=2 y=1
x=2 y=-1
x=-1 y=2
x=0 y=-3
x=-3 y=0
x(x-1)-(x2-y)=-3,求x2-y2-2xy的值
x(x-1)-(x2-y)=-3,求x2+y2-2xy的值
已知x(x-1)-(x2-y)=-3,求x2+y2-2xy的值
已知x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以(x2+y2/x
已知x2-2xy=3y2且x,y为非零实数 求:(1)2x+y/x-2y (2)2x2-3y2+5xy/x2+xy+y2
先化简再求值:-(x2+y2)+[-3xy-(x2-y2)],其中x=-1,y=2.
先化简再求值2(x2-xy)-3(2x2-3xy)-2[x2-(2x2-xy+y2)],其中x=-1,y=-2.
设x,y为实数,且x2+xy+y2=1,求x2-xy+y2的值的范围
求方程x+y/x2-xy+y2=3/7的整数解
已知x-y+1,X2+Y2=25 求(x+y)2和x2-xy+y2的值
2x2+xy-3y2+x+4y-1因式分解
当x=2,y=1 求{(x2+y2)/(x2-2xy+y2)+(2)/xy÷[(1/x)-(1/y)]}÷(x+y)的值