1、若m,n满足m+n=-2005,mn=7,求(m²+2004m+6)(n²+2006n+8)的值
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/19 05:44:05
1、若m,n满足m+n=-2005,mn=7,求(m²+2004m+6)(n²+2006n+8)的值.
2、设x(x-1)-(x²-y)=-2,求1/2(x²+y²)-xy的值.
2、设x(x-1)-(x²-y)=-2,求1/2(x²+y²)-xy的值.
(m²+2004m+6)(n²+2006n+8)
=[m(m+2004)+6][n(n+2006)+8]
=[m(-n-1)+6][n(-m+1)+8]
=(-mn+6-m)(-mn+n+8)
=-(1+m)(n+1)
=-(mn+m+n+1)
=-(7-2005+1)
=1997
x(x-1)-(x²-y)=-2
x^2-x-x^2+y=-2
x-y=2
(x²+y²)/2-xy
=1/2(x²+y²-2xy)
=(x-y)^2/2
=2^2/2
=4/2
=2
=[m(m+2004)+6][n(n+2006)+8]
=[m(-n-1)+6][n(-m+1)+8]
=(-mn+6-m)(-mn+n+8)
=-(1+m)(n+1)
=-(mn+m+n+1)
=-(7-2005+1)
=1997
x(x-1)-(x²-y)=-2
x^2-x-x^2+y=-2
x-y=2
(x²+y²)/2-xy
=1/2(x²+y²-2xy)
=(x-y)^2/2
=2^2/2
=4/2
=2
若mn互为倒数,且m+n+2005=0求(m²+2006m+1)×(n²+2006n+1)的值
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