✔ 最佳答案
1.學生若干人,每 8 人一數剩 5 個、每 10 人一數剩7 個、每 12 人一數
剩 9 個,若學生不超過 700 人,則這些學生人數最多有幾人?
Sol
x/8=______.....5
x/10=______...7
x/12=______...9
=>
(x+3)/8=_____
(x+3)/10=____
(x+3)/12=____
[8,10,12]|(x+3)
120|(x+3)
x+3=120p
120p<=700
p<=5.83
p<=5
120*5-3=597
最多有597人
2.已知 x,y 都為正整數,x被7除餘6,y被7除餘5,求(1)xy被7除餘多少?
(2)x+2y被7除餘多少?
Sol
x=7p+6
y=7q+5
(1) xy=(7p+6)(7q+5)=49pq+35p+42q+30
30/7=4……….2
餘2
(2)x+2y=(7p+6)+2(7q+5)=7p+14q+16
16/7=2…….2
餘2
3.已知a^3=5 求(a+1)(a-1)(a^2-a+1)(a^2+a+1)=?
Sol
(a+1)(a-1)(a^2-a+1)(a^2+a+1)
=(a+1)(a^2-a+1)(a-1)(a^2+a+1)
=(a^3+1)(a^3-1)
=6*4
=24
4.因式分解各式:
(1) (x-y)^2(b-c)-(y-x)(c-b)
=(x-y)^2(b-c)-(x-y)(b-c)
=(x-y)(b-c)[(x-y)-1]
=(x-y)(b-x)(x-y-1
(2) (x-y)(a-b-c)+(x+y)(c+b-a)
=(x-y)(a-b-c)-(x+y)(a-b-c)
=(a-b-c)(x-y-x-y)
=-2(a-b-c)y
5.因式分解(x+1)(x+3)(x+4)(x+6)-280
=(x+1)(x+6)(x+3)(x+4)-280
=(x^2+7x+6)(x^2+7x+12)-280
=(x^2+7x)^2+18(x^2+7x)+72-280
=(x^2+7x)^2+18(x^2+7x)-208
=(x^2+7x-8)(x^2+7x+26)
=(x-1)(x+8)(x^2+7x+26)