方程组(xy+x)/(x+y+1)=2(xz+2x)/(z+y+2)=3(y+1)(z+2)/(z+y+3)=4解出xyzshangmiancuole方程组(xy+x)/(x+y+1)=2(xz+2x)/(z+x+2)=3{(y+1)(z+2)}/(z+y+3)=4

Question

方程组(xy+x)/(x+y+1)=2 (xz+2x)/(z+y+2)=3 (y+1)(z+2)/(z+y+3)=4
解出x y z
shang mian cuo le
方程组(xy+x)/(x+y+1)=2 (xz+2x)/(z+x+2)=3 {(y+1)(z+2)}/(z+y+3)=4

Answer 1

(xy+x)/(x+y+1)=4
倒数一下,推出：
(x+y+1)/[(y+1)*x]=1/4
就是：
1/x+1/(y+1)=1/4
同样方法,后面两个可以化为：
1/(z+2)+1/x=1/2
1/(z+2)+1/(y+1)=3/4
由这两个相减得：
1/x-1/(y+1)=-1/2
再与第一个联立,解得：
x=-8
y=5/3
z=-2/5
ps:要累死人啊,楼主～