1)在3個連續正整數中，最小的是M，若其餘兩個的積是M+10，找出那三個連續數。
2)如果一個三角形的底增加10%，高減少了10%，求新三角形面積佔原來的三角形面積的百分比。
3)求2的2次方+4的2次方+6的2次方+........+200的2次方的值。
温馨提示:1的2次方+2的2次方+3的2次方+.......+n的2次方=6分之n(n+1)(2n+1)
4)求1+(1+2)+(1+2+3)+...........+(1+2+3+..........+100)的值。温馨提示:1=2分之1(2)，1+2=2分之2(3)，1x2+2x3+3x4+....+n(n+1)=3分之n(n+1)(n+2)。唔該哂~10分
數學難題!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!10分
2008-02-01 1:54 am
回答 (4)
2008-02-01 2:56 am
1) (m+1)(m+2)=m^2+3m+2=m+10
m^2+3m=m+8
m^2+2m=8
m(m+2)=8
(m+1)^2-1=8
m+1=3
m=2
that will be2,3,4
2)(1+10%)(1-10%)=110%x90%=1.1x0.9=0.99
0.99x100%=99%
3)2的2次方+4的2次方+6的2次方+........+200的2次方
=100x101x201/6=338350
4)338350
m^2+3m=m+8
m^2+2m=8
m(m+2)=8
(m+1)^2-1=8
m+1=3
m=2
that will be2,3,4
2)(1+10%)(1-10%)=110%x90%=1.1x0.9=0.99
0.99x100%=99%
3)2的2次方+4的2次方+6的2次方+........+200的2次方
=100x101x201/6=338350
4)338350
2008-02-01 2:49 am
1.
(m+1) (m+2) = m+10
m^2 +3m +2 = m+10
m^2 +2m -8=0
=> m= -4 or 2
三個連續數是 2,3,4
2.
let triangle height = h
base = b
area = h*b / 2
when h' = (h * 0.9), b' = (b*1.1)
new area = h' b'/2 = (h*b * 0.99) / 2
ratio of two triangle is 1 : 0.99
3.
P = 2^2 + 4^2 + 6^2 + ... + 200^2
= 2*2 + 4*4 + 6*6 +... + 200*200
=>
P / 4 = (2*2/2*2) + (4*4 /2*2) + (6*6/ 2*2) + ... + (200*200 / 2*2)
= 1*1 + 2*2 + 3*3 + ... + 100 * 100
= 1/6 * 100 * (100+1) * (200+1)
= 338350
=> P = 338350 *4 = 1353400
4.
R = 1+(1+2)+(1+2+3)+...........+(1+2+3+..........+100)
(from the 温馨提示)
1 = 2/2
1+2 = 2*3/2
1+2+3 = 3*4/2
1+2+3+4 = 4*5/2
...
= 2/2 + 2*3 / 2 + (3*4)/2 + ... + 100*101 /2
for R/2 = 1*2 + 2*3+ ... 100*101 = 1/3 *100*(100+1) *(100+2)
=343400
=> R = 686800
2008-01-31 18:52:01 補充：
太急post出泥第四題應該係= 2/2 + 2*3 / 2 + (3*4)/2 + ... + 100*101 /2for R*2 = 1*2 + 2*3+ ... 100*101 = 1/3 *100*(100+1) *(100+2) =343400 R = 171700
(m+1) (m+2) = m+10
m^2 +3m +2 = m+10
m^2 +2m -8=0
=> m= -4 or 2
三個連續數是 2,3,4
2.
let triangle height = h
base = b
area = h*b / 2
when h' = (h * 0.9), b' = (b*1.1)
new area = h' b'/2 = (h*b * 0.99) / 2
ratio of two triangle is 1 : 0.99
3.
P = 2^2 + 4^2 + 6^2 + ... + 200^2
= 2*2 + 4*4 + 6*6 +... + 200*200
=>
P / 4 = (2*2/2*2) + (4*4 /2*2) + (6*6/ 2*2) + ... + (200*200 / 2*2)
= 1*1 + 2*2 + 3*3 + ... + 100 * 100
= 1/6 * 100 * (100+1) * (200+1)
= 338350
=> P = 338350 *4 = 1353400
4.
R = 1+(1+2)+(1+2+3)+...........+(1+2+3+..........+100)
(from the 温馨提示)
1 = 2/2
1+2 = 2*3/2
1+2+3 = 3*4/2
1+2+3+4 = 4*5/2
...
= 2/2 + 2*3 / 2 + (3*4)/2 + ... + 100*101 /2
for R/2 = 1*2 + 2*3+ ... 100*101 = 1/3 *100*(100+1) *(100+2)
=343400
=> R = 686800
2008-01-31 18:52:01 補充：
太急post出泥第四題應該係= 2/2 + 2*3 / 2 + (3*4)/2 + ... + 100*101 /2for R*2 = 1*2 + 2*3+ ... 100*101 = 1/3 *100*(100+1) *(100+2) =343400 R = 171700
2008-02-01 2:24 am
1)(M+1)(M+2)=M+10
M^2+3M+2=M+10
M^2+2M-8=0
(M+4)(M-2)=0
M=2 或 M=-4(捨去)
2)設三角形的底增是x，而高是y。
x(1+10%) X y(1-10%)
-------------------------------------------X100%
xy
xy(110% X 90%)
=----------------------------------- X100%
xy
=99%
M^2+3M+2=M+10
M^2+2M-8=0
(M+4)(M-2)=0
M=2 或 M=-4(捨去)
2)設三角形的底增是x，而高是y。
x(1+10%) X y(1-10%)
-------------------------------------------X100%
xy
xy(110% X 90%)
=----------------------------------- X100%
xy
=99%
2008-02-01 2:01 am
2) 100%
收錄日期: 2021-04-13 15:03:56
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080131000051KK01684