问题
解答题
已知x=y=11,求(xy-1)2+(x+y-2)(x+y-2xy)的值.
答案
设x+y=m,xy=n.
原式=(n-1)2+(m-2)(m-2n)
=(n-1)2+m2-2m-2mn+4n
=n2-2n+1+4n-2m-2mn+m2
=(n+1)2-2m(n+1)+m2
=(n+1-m)2
=(11×11+1-22)2
=(121+1-22)2
=1002
=10000.
已知x=y=11,求(xy-1)2+(x+y-2)(x+y-2xy)的值.
设x+y=m,xy=n.
原式=(n-1)2+(m-2)(m-2n)
=(n-1)2+m2-2m-2mn+4n
=n2-2n+1+4n-2m-2mn+m2
=(n+1)2-2m(n+1)+m2
=(n+1-m)2
=(11×11+1-22)2
=(121+1-22)2
=1002
=10000.