一道三元二次方程组数学题x^2=6+(z-y)^2y^2=2+(z-x)^2z^2=3=(x-y)^2
2019-05-28
一道三元二次方程组数学题
x^2=6+(z-y)^2
y^2=2+(z-x)^2
z^2=3=(x-y)^2
优质解答
x^2=6+(z-y)^2变换为x^2-(z-y)^2=6,其他两式同理.再根据平方差公式打开,
(x+y-z)(x-y+z)=6,(-x+y+z)(x+y-z)=2,(-x+y+z)(x-y+z)=3,为了方便,
把(x+y-z)设为A,(x-y+z)设为B,(-x+y+z)设为C.则A*B=6,AC=2,BC=3,可导出C方=1,再根据xyz与ABC的关系,应该可以求出结果.
x^2=6+(z-y)^2变换为x^2-(z-y)^2=6,其他两式同理.再根据平方差公式打开,
(x+y-z)(x-y+z)=6,(-x+y+z)(x+y-z)=2,(-x+y+z)(x-y+z)=3,为了方便,
把(x+y-z)设为A,(x-y+z)设为B,(-x+y+z)设为C.则A*B=6,AC=2,BC=3,可导出C方=1,再根据xyz与ABC的关系,应该可以求出结果.