(X+2y)(x+2z)=-16(y+2z)(y+2x)=8(z+2x)(z+2y)=-7解方程组.

Question

(X+2y)(x+2z)=-16 (y+2z)(y+2x)=8 (z+2x)(z+2y)=-7解方程组.

Answer 1

x+2y=(x+y+z)+(y-z),x+2z=(x+y+z)-(y-z),故第一个方程可变形为：
(x+y+z)²-(y-z)²=-16 ············①
同理第二、三个方程可变形为：
(x+y+z)²-(z-x)²=8 ············②
(x+y+z)²-(x-y)²=-7 ············③
②-①得：(y-z)²-(z-x)²=24 ············④②-③得：(x-y)²-(z-x)²=15 ············⑤
令y-z=a,z-x=b,由于(x-y)+(y-z)+(z-x)=0,故x-y=-(y-z)-(z-x)=-(a+b)
于是方程④、⑤变为：
a²-b²=24 ············⑥
(a+b)²-b²=15············⑦
⑦×8-⑥×5得：8(a+b)²-8b²=5(a²-b²),整理后为(3a+b)(a+5b)=0,a=-b/3或a=-5b
将a=-b/3代入⑥可知此时无解,舍弃.将a=-5b代入⑥可解得两组
a=5、b=-1或a=-5、b=1
（1）a=5、b=-1时
此时y-z=5,z-x=-1,将y-z=5代入①可得x+y+z=3或x+y+z=-3
当x+y+z=3时,将其与y-z=5、z-x=-1联立,解得x=0,y=4,z=-1
当x+y+z=-3时,将其与y-z=5、z-x=-1联立,解得x=-2,y=2,z=-3
（2）a=-5、b=1时
此时y-z=-5,z-x=1,将y-z=-5代入①同样可得x+y+z=3或x+y+z=-3
当x+y+z=3时,将其与y-z=-5、z-x=1联立,解得x=2,y=-2,z=3
当x+y+z=-3时,将其与y-z=-5、z-x=1联立,解得x=0,y=-4,z=1
综上,(x,y,z)有四组(0,4,-1)、(-2,2,-3)、(2,-2,3)、(0,-4,1)