作业帮已知ax+by=7,ax^2+by^2=49,ax^3+by^3=133,ax^4+by^4=406.求：2001（x+
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已知ax+by=7,ax^2+by^2=49,ax^3+by^3=133,ax^4+by^4=406.求：2001（x+y）+6xy-17/2（a+b）的值
设
x+y = m
xy = n
a + b = k
(x+y)(ax+by) = ax^2+by^2+(a+b)xy =>7m = 49+kn (1)
(x+y)(ax^2+by^2) = ax^3+by^3+(ax+by)xy => 49m = 133 + 7n (2)
(x+y)(ax^3+by^3) = ax^4+by^4+(ax^2+by^2)xy => 133m = 406 + 49n (3)
由(2)(3)得
m = 5/2
n = -3/2
代入(1)得k = 21
所以2001(x+y) + 6xy - 17/2(a+b) = 2001* 5/2 - 6*(3/2) - 17/2*21 = 4815
x+y = m
xy = n
a + b = k
(x+y)(ax+by) = ax^2+by^2+(a+b)xy =>7m = 49+kn (1)
(x+y)(ax^2+by^2) = ax^3+by^3+(ax+by)xy => 49m = 133 + 7n (2)
(x+y)(ax^3+by^3) = ax^4+by^4+(ax^2+by^2)xy => 133m = 406 + 49n (3)
由(2)(3)得
m = 5/2
n = -3/2
代入(1)得k = 21
所以2001(x+y) + 6xy - 17/2(a+b) = 2001* 5/2 - 6*(3/2) - 17/2*21 = 4815
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