已知(x+y)^2=1,(x-y)^2=11.,x^2+y^2=?xy=?
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已知(x+y)^2=1,(x-y)^2=11.,x^2+y^2=?xy=?
已知(x+y)^2=1,(x-y)^2=11.,x^2+y^2=?xy=?
已知(x+y)^2=1,(x-y)^2=11.,x^2+y^2=?xy=?
即:
x²+2xy+y²=1 ①
x²-2xy+y²=11 ②
①+②得:2x²+2y²=12,则:x²+y²=6
①-②得:4xy=-10,则:xy=-2.5
(x+y)²-(x-y)²
=(x+y+x-y)(x+y-x+y)
=2x*2y
=4xy=1-11=-10
所以xy=-5/2
因为(x+y)²=x²+y²+2xy=1
所以x²+y²=1-2xy=1+5=6
赞成楼上。。。
(x+y)^2=1,
(x-y)^2=11
两式相加得:2(x^2+y^2)=12, 故有x^2+y^2=6
两式相减得:4xy=-10, 故有xy=-5/2
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