方程组xy+yz+zx=1(1) yz+zt+ty=1(2) zt+tx+xz=1(3) tx+xy+yt=1(4)
的解是x1=33 y1=33 z1=33 t1=33
,x2=-33 y2=-33 z2=-33 t2=-33
x1=33 y1=33 z1=33 t1=33
,x2=-33 y2=-33 z2=-33 t2=-33
.
xy + yz + zx = 1 ( 1 ) |
yz + zt + ty = 1 ( 2 ) |
zt + tx + xz = 1 ( 3 ) |
tx + xy + yt = 1 ( 4 ) |
x 1 = 3 3 |
y 1 = 3 3 |
z 1 = 3 3 |
t 1 = 3 3 |
,
x 2 = - 3 3 |
y 2 = - 3 3 |
z 2 = - 3 3 |
t 2 = - 3 3 |
x 1 = 3 3 |
y 1 = 3 3 |
z 1 = 3 3 |
t 1 = 3 3 |
,
x 2 = - 3 3 |
y 2 = - 3 3 |
z 2 = - 3 3 |
t 2 = - 3 3 |
【考点】高次方程.
【答案】
x 1 = 3 3 |
y 1 = 3 3 |
z 1 = 3 3 |
t 1 = 3 3 |
,
x 2 = - 3 3 |
y 2 = - 3 3 |
z 2 = - 3 3 |
t 2 = - 3 3 |
【解答】
【点评】
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发布:2025/5/26 21:30:2组卷:150引用:1难度:0.3
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