根
练习6.2
问题1:求下列数的平方。
(我)32
答:322= 32 * 32 = 1024
但上述方法计算难度较大。计算这些值的方法如下:
因为，32可以写成(30+2)
因此,322= (30 + 2)2= (30 + 2) (30 + 2)
= 30(30+2)+2(30+2) = 302+ 30 × 2 + 2 × 30 + 22
= 900 + 60 + 60 + 4 = 1024
答案:1024
(2) 35
答:(35）2= (30 + 5)2= (30 + 5) (30 + 5)
= 30(30+5)+5(30+5) = 302+ 30 × 5 + 5 × 30 + 52
= 900 + 150 + 150 + 25 = 1225
因此，Answer = 1225
(3) 86
答:862= (80 + 6)2= (80 + 6)(80 + 6)
= 802+ 80 × 6 + 6 × 80 + 62
= 6400 + 480 + 480 + 36 = 7396
因此，答案是7396
(四)93
答:932= (90 + 3)2= (90 + 3) (90 + 3)
= 90 (90 + 3) + 3 (90 + 3) = 902+ 90 × 3 + 3 × 90 + 32
= 8100 + 270 + 270 + 9 = 8649
因此，答案是:8649
(v) 71
答:712= (70 + 1)2= (70 + 1) (70 + 1)
= 70 (70 + 1) + 1 (70 + 1) = 702+ 70 × 1 + 1 × 70 + 1 × 1
= 4900 + 70 + 70 + 1 = 4900 + 140 + 1 = 5040 + 1 = 5041
因此，答案是5041
(vi) 46
答:462= (40 + 6)2(40 + 6) (40 + 6)
= 40 (40 + 6) + 6 (40 + 6) = 402+ 40 × 6 + 6 × 40 + 62
= 1600 + 240 + 240 + 36 = 1600 + 480 + 36 = 2080 + 36 = 2116
因此，答案是2116
问题2:写出一个毕达哥拉斯三联，其中一个成员是:
(我)6
答:我们知道2m m2+ 1和m2- 1对任意数m > 1构成毕达哥拉斯三连式。
我们假设2m = 6
因此，m = 3
,米2+ 1 = 32+ 1= 9 + 1= 10
,米2- 1 = 32- 1 = 9 - 1 = 8
测试:62+ 82= 36 + 64 = 100 = 102
因此，三联体是6,8,10答案
(2) 14
答:假设2m = 14，则m = 7
现在,米2+ 1 = 72+ 1 = 49 + 1 = 50
,米2- 1 = 72- 1 = 49 - 1 = 48
测试:142+ 482= 196 + 1304 = 2500 = 502
因此，三联体是14、48和50
(3) 16
答:我们假设2m = 16，那么m = 8
现在,米2+ 1 = 82+ 1 = 64 + 1 = 65
,米2- 1 = 82- 1 = 64 - 1 = 63
测试:162+ 632= 256 + 3969 = 4225 = 652
因此，三联体是16,63和65答案
(iv) 18
答:我们假设2m = 18，因此m = 9
现在,米2+ 1 = 92+ 1 = 81 + 1 = 82
,米2- 1 = 92- 1 = 81 - 1 = 80
测试:182+ 802= 324 + 6400 = 6724 = 822