xx+2xy+yy=1 xx-2xy+yy=49 求xx+yy,xy.

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xx+2xy+yy=1 xx-2xy+yy=49 求xx+yy,xy.
xx+2xy+yy=1 xx-2xy+yy=49 求xx+yy,xy.

xx+2xy+yy=1 xx-2xy+yy=49 求xx+yy,xy.
x²+y²
=[(x²+2xy+y²)+(x²-2xy+y²)]÷2
=(1+49)÷2
=50÷2
=25
xy
=[(x²+2xy+y²)-(x²-2xy+y²)]÷4
=(1-49)÷4
=-48÷4
=-12

xx+yy=（1+49）÷2=25
xy=(1-49)÷4=-12

两式相加除以2，得xx+yy=25
两式相减除以4，得xy=-12

xx+yy=25.xy=-24

xx+yy=25
xy=-12

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