若实数x,y,z满足x+1⼀y=4,y+1⼀z=1,z+1⼀x=7⼀3,求x,y,z的值

2024-12-27 12:28:50
推荐回答(2个)
回答1:

x+1/y=4..........(1)
y+1/z=1.........(2)
z+1/x=7/3....(3)
(1): 1/y=4-x
y=1/(4-x).......(4)
代入(2):
1/(4-x)+1/z=1
1/z=1-1/(4-x)
=(3-x)/(4-x)
z=(4-x)/(3-x)........(5)
(5)代入(3):
(4-x)/(3-x)+1/x=7/3
[x(4-x)+(3-x)]/[x(3-x)]=7/3
3(4x-x^2+3-x)=7x(3-x)
9x-3x^2+9=21x-7x^2
4x^2-12x+9=0
(2x-3)^2=0
2x-3=0
x=3/2
代入(4):
y=1/(4-3/2)
=1/[8-3)/2]
=2/5
代入(2):
2/5+1/z=1
1/z=1-2/5
=3/5
z=5/3
综上,x=3/2、y=2/5、z=5/3

回答2:

三式相加:x+y+z+1/x+1/y+1/z=22/3
三式相乘:xyz+y+x+1/z+z+1/x+1/y+1/xyz=28/3
将1式代入2式
得到xyz+22/3+1/xyz=28/3
即:xyz+1/xyz=2.所以xyz=1