x+y+z=0 求x(1/y+1/z)+y(1/x+1z)+z(1/x+1/y)=?

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x+y+z=0 求x(1/y+1/z)+y(1/x+1z)+z(1/x+1/y)=?

x+y+z=0 求x(1/y+1/z)+y(1/x+1z)+z(1/x+1/y)=?
x+y+z=0 求x(1/y+1/z)+y(1/x+1z)+z(1/x+1/y)=?

x+y+z=0 求x(1/y+1/z)+y(1/x+1z)+z(1/x+1/y)=?
x+y+z=0
x=-y-z
y=-x-z
z=-x-y
带入x(1/y+1/z)+y(1/x+1z)+z(1/x+1/y)
=-(y+z)(1/y+1/z)-(x+z)(1/x+1z)-(x+y)(1/x+1/y)
=-6-[(y+z)/x+(x+z)/y+(x+y)/z]
继续带入
-x=y+z
-y=x+z
-z=x+y
原式=-6+3=-3

x(1/y+1/z)+y(1/x+1z)+z(1/x+1/y)
=x/y+x/z+y/x+y/z+z/x+z/y
=(x+y)/z+(y+z)/x+(x+z)/y
=(0-z)/z+(0-x)/x+(0-y)/y
=-1-1-1
=-3

因为x+y+z=0
所以
y+z=-x
x+z=-y
x+y=-z
x(1/y+1/z)+y(1/x+1z)+z(1/x+1/y)
=x/y+x/z+y/x+y/z+z/x+z/y
=y/x+z/x+x/y+z/y+x/z+y/z
=(y+z)/x+(x+z)/y+(x+y)/z
=(-x)/x+(-y)/y+(-z)/Z
=-3

这题关键在于等式变形。你要求的式子先变形，把各项展开，再把同分母相加一下，就变成：（Z+Y)/X+(X+Z)/Y+(x+y)/z =? 你把给的等式分别移项一下便可以得到与分母异号，式子变为：-Z/Z + -Y/Y + -z/z = -3

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