Question

因式分解；（x-y)^2-4(x-y-1)=多少

2^48能被60与70之间的两个整数整除，找出这两个整数，并说明理由
这两题怎么做

Answer 1

第一题：
（x-y)²-4(x-y-1)
=（x-y)²-4(x-y)+4
=[（x-y)-2]²
=(x-y-2)²

第二题：题目错了，应该是：2^48-1可以被60至70之间的两个整数整除，这两个整数是多少
2^48-1
=(2^24+1)(2^24-1)
=(2^24+1)(2^12+1)(2^12-1)
=(2^24+1)(2^12+1)(2^6+1)(2^6-1)
=(2^24+1)(2^12+1)*65*63

所以两个数是65和63

Answer 2

（x-y)^2-4(x-y-1)
=(x-y)²-4(x-y)+4
=(x-y-2)²

2^48-1
=(2^24+1)(2^24-1)
=(2^24+1)(2^12+1)(2^12-1)
=(2^24+1)(2^12+1)(2^6+1)(2^6-1)
=(2^24+1)(2^12+1)*65*63
所以两个数是65和63

追问

第一题为什么+4

追答

（x-y)^2-4(x-y-1)
=（x-y）²-4(x-y)-4*(-1)

=(x-y)²-4(x-y)+4

Answer 3

（x-y)^2-4(x-y-1)
=(x-y)²-4（x-y)+4
=(x-y-2)²

2^48-1
=(2^24-1)(2^24+1)
=(2^12-1)(2^12+1)(2^24+1)
=(2^6-1)(2^6+1)(2^12+1)(2^24+1)
=63*65*(2^12+1)(2^24+1)
∴2^48能被60与70之间的63和65两个整数整除

Answer 4

（x-y)^2-4(x-y-1)=（x-y)^2-4(x-y）+4=（x-y-2)²

追问

还有下面一题

追答

2^48-1
=（2^24）²-1
=(2^24+1)(2^24-1)
=(2^24+1)(2^12+1)(2^12-1)
=(2^24+1)(2^12+1)(2^6+1)(2^6-1)
=(2^24+1)(2^12+1)*65*63

两个数是65和63