Discussion:
a problem
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托斯卡尼艷陽下
2009-08-09 14:54:11 UTC
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a是個正實數, b為a的小數部分, 0<= b <1,

(Ex: a=1.5, 則 b=0.5)

已知 a^2+b^2=6, 求a=?


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(short)(-15074)
2009-08-09 23:32:43 UTC
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Post by 托斯卡尼艷陽下
a是個正實數, b為a的小數部分, 0<= b <1,
(Ex: a=1.5, 則 b=0.5)
已知 a^2+b^2=6, 求a=?
令 a-b=整數n

則知 (n+b)^2+b^2=6

n^2+2nb+2b^2=6

因此 n 有幾個可能:

(1) n=2: 此時 2b^2+4b=2 => b^2+2b-1=0 => b=-1±√2 僅-1+√2合條件

故 a=n+b=2+(-1+√2)=1+√2

(2) n=1: 2b^2+2b=5 => b=(-1±√11)/2 但因√11>3, 故-1+√11>2 不合

(3) n=0: 2b^2=6 b=±√3 顯然不合

(4) n=-1: 2b^2-2b=5 => b=(1±√11)/2 同樣不合

(5) n=-2: 2b^2-4b=2 => b=1±√2 依然不合

故僅有 a=1+√2 一解

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髒筆
2009-08-10 07:15:45 UTC
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Post by (short)(-15074)
Post by 托斯卡尼艷陽下
a是個正實數, b為a的小數部分, 0<= b <1,
(Ex: a=1.5, 則 b=0.5)
已知 a^2+b^2=6, 求a=?
令 a-b=整數n
則知 (n+b)^2+b^2=6
n^2+2nb+2b^2=6
只有一種可能

0≦b<1 => 0≦b^2<1 => 5<a^2≦6 => n=2or-2 (題目有說a是個正實數)
Post by (short)(-15074)
(1) n=2: 此時 2b^2+4b=2 => b^2+2b-1=0 => b=-1±√2 僅-1+√2合條件
故 a=n+b=2+(-1+√2)=1+√2
(2) n=1: 2b^2+2b=5 => b=(-1±√11)/2 但因√11>3, 故-1+√11>2 不合
(3) n=0: 2b^2=6 b=±√3 顯然不合
(4) n=-1: 2b^2-2b=5 => b=(1±√11)/2 同樣不合
(5) n=-2: 2b^2-4b=2 => b=1±√2 依然不合
故僅有 a=1+√2 一解
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髒筆
2009-08-10 07:50:40 UTC
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m,n是介於1到100之間的兩正整數, m>n,
令f(m,n)= |m^(1/2)+n^(1/2)-17.1|
試求 f(m,n) 的最小值, 且請問最小值發生時的(m,n)=?
用程式跑的結果

f(m,n)的最小值是發生在 f(84,63) = | 0.0024053231 |

但是如果稍稍超過範圍馬上有更好的最小值
f(102,49) = | -0.0004950616 |

擴大範圍到正整數時,前三名為

 冠軍 f(154,22) = | 0.0000894058 |
f(133,31) = | 0.0003269575 |
f(228,4) = | -0.0003311294 |

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