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已知xy=myz=n

2021-10-18 8 56jieju

由x-y=m,y-z=n

两式相加,得

x-z=m+n

所以

x^2+y^2+z^2-xy-yz-zx

=(2x^2+2y^2+2z^2-2xy-2yz-2zx)/2

=[(x^2+y^2-2xy)+(x^2+z^2-2zx)+(y^2+z^2-2zy)]/2

=[(x-y)^2+(x-z)^2+(y-z)^2]/2

=[m^2+n^2+(m+n)^2]/2

=m^2+n^2+mn

因为x-y=m,y-z=n两式相加 ,得

x-z=m+n

原式=(2x^2+2y^2+2z^2-2xy-2yz-2zx)/2

=〔(x^2-2xy+y^2)+(y^2--2yz+z^2)+(x^2-2zx+z^2)〕/2

=〔(x-y)^2+(y-z)^2+(x-z)^2〕/2

=〔m^2+n^2+(m+n)^2〕/2

=m^2+mn+n^2

x-y = m, y-z = n ===> x-z = m+n

因此:

x^2+y^2+z^2-xy-yz-zx

= [(x-y)^2 + (y-z)^2 + (z-x)^2]/2

= [m^2 + n^2 + (m+n)^2]/2

= m^2 + n^2 + mn

因为x-y=m,y-z=n ,所以 x-z=m+n

x^2+y^2+z^2-xy-yz-zx=(x^2-2xy+y^2)/2+(y^2-2yz+z^2)/2+(x^2-2zx+z^2)/2

=1/2*[(x-y)^2+(y-z)^2+(x-z)^2]

=1/2*[m^2+n^2+(m+n)^2]

相关标签: # xy # zx # yz

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