已知x+y+z=2,xy+yz+zx=-5,求x²+y²+z²的值
问题描述：
已知x+y+z=2,xy+yz+zx=-5,求x²+y²+z²的值
答
(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz
x²+y²+z²=(x+y+z)^2-2(xy+yz+zx)=4+10=14
答
(x＋y＋z)²=x²＋y²＋z²＋2xy＋2yz＋2xz
4=(x²＋y²＋z²)－10
x²＋y²＋z²=14
答
(x+y+z)^2=4
x^2+y^2+z^2+2xy+2xz+2yz=4
x^2+y^2+z^2+2(-5)=4
x^2+y^2+z^2=14