已知xy/(x+y)=a,xz/(x+z)=b,yz/(y+z)=c,且a.b.c不等于0,求x的值

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已知xy/(x+y)=a,xz/(x+z)=b,yz/(y+z)=c,且a.b.c不等于0,求x的值已知xy/(x+y)=a,xz/(x+z)=b,yz/(y+z)=c,且a.b.c不等于0,求x的值

已知xy/(x+y)=a,xz/(x+z)=b,yz/(y+z)=c,且a.b.c不等于0,求x的值
已知xy/(x+y)=a,xz/(x+z)=b,yz/(y+z)=c,且a.b.c不等于0,求x的值

已知xy/(x+y)=a,xz/(x+z)=b,yz/(y+z)=c,且a.b.c不等于0,求x的值
a≠0,xy≠0 x≠0且y≠0;同理,b≠0,x≠0,z≠0
综上,得x,y,z≠0
xy/(x+y)=a
(x+y)/(xy)=1/a
1/x+1/y=1/a (1)
同理
1/x+1/z=1/b (2)
1/y+1/z=1/c (3)
[(1)+(2)+(3)]/2
1/x+1/y+1/z=(1/a+1/b+1/c)/2 (4)
(4)-(3)
1/x=(1/a+1/b+1/c)/2 -1/c=1/(2a)+1/(2b)-1/(2c)=(bc+ac-ab)(2abc)
x=2abc/(bc+ac-ab)
还可以求出y、z
(4)-(2)
1/y=(1/a+1/b+1/c)/2 -1/b=1/(2a)-1/(2b)+1/(2c)=(bc-ac+ab)/(2abc)
y=2abc/(bc-ac+ab)
(4)-(1)
1/z=(1/a+1/b+1/c)/2-1/a=1/(2b)+1/(2c)-1/(2a)=(ac+ab-bc)/(2abc)
z=(2abc)/(ab+ac-bc)

三个方程分别两边取倒数得
1/x + 1/y=1/a①
1/x + 1/z=1/b②
1/y + 1/z=1/c③
这三个式子相加得
2(1/x + 1/y +1/z)=1/a + 1/b + 1/c
则1/x + 1/y +1/z=1/2(1/a + 1/b + 1/c)④
④-③得1/x=1/2(1/a + 1/b - 1/c)
则x=2/(1/a + 1/b - 1/c)