Question

xy／x+y＝1,yz／y＋z＝2,xz／x＋z＝3求x,y

Answer 1

xy/(x+y)=1,取倒数

(x+y)/xy=1

x/xy+y/xy=1

1/y+1/x=1.1

yz/(y+z)=2,取倒数

(y+z)/yz=1/2

y/yz+z/yz=1/2

1/z+1/y=1/2.2

xz/(x+z)=3取倒数

(x+z)/xz=1/3

x/xz+z/xz=1/3

1/z+1/x=1/3.3

1式+2式+3式得

2(1/x+1/y+1/z)=1+1/2+1/3

2(1/x+1/y+1/z)=11/6

1/x+1/y+1/z=11/12.4

将3式代入4式得

1/y+1/3=11/12

1/y=7/12

y=12/7

将2式代入4式得

1/x+1/2=11/12

1/x=5/12

x=12/5