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已知x+y+z=4,x*x+y*y+z*z=2,求xy+yz+zx的值
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已知x+y+z=4,x*x+y*y+z*z=2,求xy+yz+zx的值
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答案和解析
（x+y+z=4）^2=x*x+y*y+z*z+2xy+2yz+2zx=16
x*x+y*y+z*z=2,则
xy+yz+zx=(（x+y+z=4）^2-x*x+y*y+z*z)/2=(16-2)/2=7
x*x+y*y+z*z=2,则
xy+yz+zx=(（x+y+z=4）^2-x*x+y*y+z*z)/2=(16-2)/2=7
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