设a1,a2,a3,a4,a5为自然数,A={a1,a2,a3,a4,a5},B={a1^2,a2^2,a3^2,a4^
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设a1,a2,a3,a4,a5为自然数,A={a1,a2,a3,a4,a5},B={a1^2,a2^2,a3^2,a4^2,a5^2},且a1小于a2小于a3小于a4小于a5,并满足A与B的交集={a1,a4},a1+a4=10,A与B的并集中各元素之和为256,求集合A
(1)
因为A与B的交集={a1,a4},所以B中含有a1和a4,a1与a4都是平方数
而a1+a4=10,故只能取 a1 = 1 ,a4 = 9
所以A = { 1,3,x,9,a5 } …………(注:可能是a2=3,也可能是a3=3)
B = { 1,9,x^2,81,a5^2 }
(2)
A与B的并集 = { 1,3,x,9,a5,x^2,81,a5^2 }
他们的和为256,故:
x + x^2 + a5 + a5^2 = 256 - (1+3+9+81) = 162
(3)
我们先从a5入手,a5>9,但 ( a5 + a5^2 ) 应小于162,所以a5只能取10、11、12
若a5 = 10,则 ( x + x^2 ) = 162 - 110 = 52,x无整数解;
若a5 = 11,则 ( x + x^2 ) = 162 - 132 = 30,可求得 x = 5;
若a5 = 12,则 ( x + x^2 ) = 162 - 156 = 6,可求得 x = 2.
综上所述,总共有2个答案:
① A = { 1,3,5,9,11 },B = { 1,9,25,81,121 }
② A = { 1,2,3,9,12 },B = { 1,4,9,81,144 }
因为A与B的交集={a1,a4},所以B中含有a1和a4,a1与a4都是平方数
而a1+a4=10,故只能取 a1 = 1 ,a4 = 9
所以A = { 1,3,x,9,a5 } …………(注:可能是a2=3,也可能是a3=3)
B = { 1,9,x^2,81,a5^2 }
(2)
A与B的并集 = { 1,3,x,9,a5,x^2,81,a5^2 }
他们的和为256,故:
x + x^2 + a5 + a5^2 = 256 - (1+3+9+81) = 162
(3)
我们先从a5入手,a5>9,但 ( a5 + a5^2 ) 应小于162,所以a5只能取10、11、12
若a5 = 10,则 ( x + x^2 ) = 162 - 110 = 52,x无整数解;
若a5 = 11,则 ( x + x^2 ) = 162 - 132 = 30,可求得 x = 5;
若a5 = 12,则 ( x + x^2 ) = 162 - 156 = 6,可求得 x = 2.
综上所述,总共有2个答案:
① A = { 1,3,5,9,11 },B = { 1,9,25,81,121 }
② A = { 1,2,3,9,12 },B = { 1,4,9,81,144 }
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