Question

数学作业 不会做急啊

81(x+y)平方-121(m+n)平方
(x平方+y平方)平方-x平方y平方
(2m-n)平方-(3m+2n)平方
两个连续奇数的平方差是8的倍数
提示 可设两个连续奇数为 2k+1 2k+3 k为正整数

Answer 1

81(x+y)²-121(m+n)²
=[9(x+y)]²-[11(m+n)]²
=(9x+9y)²-(11m+11n)²
=[(9x+9y)+(11m+11n)][(9x+9y)-(11m+11n)]
=(9x+9y+11m+11n)(9x+9y-11m-11n)

(x²+y²)²-x²y²
=(x²+y²+xy)(x²+y²-xy)

(2m-n)²-(3m+2n)²
=[(2m-n)+(3m+2n)][(2m-n)-(3m+2n)]
=(2m-n+3m+2n)(2m-n-3m-2n)
=-(5m+n)(m+3n)

两个连续奇数的平方差是8的倍数
(2k+3)²-(2k+1)²
=[(2k+3)+(2k+1)][(2k+3)-(2k+1)]
=2(4k+4)
=8(k+1)
所以是8的倍数