Question

- -问几道数学题,要过程,急~

1.简便计算:(3X+2Y)²(3X-2Y)²
2:计算:(2X+6)²-2(3-X)²
3.填空:X²+10X+( )=( )²
4.试说明两个连续奇数的平方差一定是8的倍数[用n表示]
5.计算:(2ń+1)(2ń+1)(2ń+1).......(2ń+1)
*题目中ń分别为2.4.8.32

Answer 1

1.
(3X+2Y)²(3X-2Y)²
=[(3x+2y)(3x-2y)]²
=(9x-4y)²
=81x²-72xy+4y²

2.
(2X+6)²-2(3-X)²
=4(x+3)²-2(x+3)²
=2(x+3)²
=4x²+24x+36

3.填空:X²+10X+(25)=(x+5)²

4.试说明两个连续奇数的平方差一定是8的倍数[用n表示]

(2n+1)²-(2n-1)²
=(2n+1+2n-1)(2n+1-2n+1)
=4n*2
=8n
所以是8的倍数

5.
(2ń+1)(2ń+1)(2ń+1).......(2ń+1)
=(2-1)(2ń+1)(2ń+1)(2ń+1).......(2ń+1)
=(2²-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)
=...
=2^64-1