已知x+y+z=0,则y²+z²-x²分之1+z²+x²-y²分之1+x²+y²-z²分之1=______.
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/06 19:01:07
![已知x+y+z=0,则y²+z²-x²分之1+z²+x²-y²分之1+x²+y²-z²分之1=______.](/uploads/image/z/8566389-45-9.jpg?t=%E5%B7%B2%E7%9F%A5x%2By%2Bz%3D0%2C%E5%88%99y%26%23178%3B%2Bz%26%23178%3B-x%26%23178%3B%E5%88%86%E4%B9%8B1%2Bz%26%23178%3B%2Bx%26%23178%3B-y%26%23178%3B%E5%88%86%E4%B9%8B1%2Bx%26%23178%3B%2By%26%23178%3B-z%26%23178%3B%E5%88%86%E4%B9%8B1%3D______.)
已知x+y+z=0,则y²+z²-x²分之1+z²+x²-y²分之1+x²+y²-z²分之1=______.
已知x+y+z=0,则y²+z²-x²分之1+z²+x²-y²分之1
+x²+y²-z²分之1=______.
已知x+y+z=0,则y²+z²-x²分之1+z²+x²-y²分之1+x²+y²-z²分之1=______.
答:
x+y+z=0
y²+z²-x²分之1+z²+x²-y²分之1+x²+y²-z²分之1
=1/(y²+z²-x²)+1/(z²+x²-y²)+1/(x²+y²-z²)
=1/[y²+z²-(-y-z)²]+1/[z²+x²-(-z-x)²]+1/[x²+y²-(-x-y)²]
=1/(-2yz)+1/(-2xz)+1/(-2xy)
=-(1/2)*(x+y+z)/(xyz)
=0