求寫下數學問題詳細計算過程3

2014-06-09 9:02 pm
1.試求出所有滿足(指數方程式) (4^x)+(6^x)=(9^x) 的x值。
都是實數。

2.方程式(x^3)-(6x^2)+3x+1=0 的三根a,b,r 都是實數。

(i) 試求出(a^3)+(b^3)+(r^3) 的值;
(ii) 試求出 ( (a^2)+1 ) ( (b^2) +1) ( (r^2)+1) 的值;

3.已知實數x,y,z 滿足(x+y+z) (y+z) (z+x) (x+y) 不= 0
且( (x^2) / (y+z) )+( (y^2) / (z+x) )+( (z^2) / (x+y) ) = 0 。試求:

(i) ( (x^2) / (y+z) )+( (y^2) / (z+x) )+( (z^2) / (x+y) ) 的值;
(ii) ( (x^2) / yz )+( (y^2) / zx )+( (z^2) / xy ) 的值。


求寫下詳細計算過程....thx...

回答 (5)

2014-06-09 11:06 pm
✔ 最佳答案

(1)

4^x + 6^x = 9^x
4^x + 6^x - 9^x = 0
(2^x)^2 + (2^x)(3^x) - (3^x)^2 = 0
[(2/3)^x]^2 + (2/3)^x - 1 = 0
(2/3)^x = (-1 + root5) / 2
x log(2/3) = log[(-1 + root 5) / 2]
x = [log(root 5 - 1) - log2] / (log 2 - log 3)


(2)

(x - a)(x - b)(x - r) = 0
x^3 - (a + b + r)x^2 + (ab + br + ra)x - abr = 0
so, a + b + r = 6, ab + br + ra = 3, abr = -1

a^3 + b^3 + r^3
= (a + b + r)^3 - 3(ab^2 + br^2 + ra^2 + ar^2 + ba^2 + rb^2) - 6abr
= (a + b + r)^3 - 3[(a + b + r)(ab + br + ra) - 3abr] - 6abr
= (a + b + r)^3 - 3(a + b + r)(ab + br + ra) + 3abr
= (6)^3 - 3(6)(3) + 3(-1) = 159

(a^2 + 1)(b^2 + 1)(r^2 + 1)
= (abr)^2 + (a^2)(b^2) + (b^2)(r^2) + (r^2)(a^2) + a^2 + b^2 + r^2 + 1
= (abr)^2 + (ab + br + ra)^2 - 2(arb^2 + bar^2 + rba^2)
+ (a + b + r)^2 - 2(ab + br + ra) + 1
= (abr)^2 + (ab + br + ra)^2 - 2abr(a + b + r) + (a + b + r)^2 - 2(ab + br + ra) + 1
= (-1)^2 + (3)^2 - 2(-1)(6) + (6)^2 - 2(3) + 1 = 53



2014-06-09 15:08:06 補充:
第三題有待修正

2014-06-09 15:09:59 補充:
原來貓大已答了,sorry
參考: knowledge
2014-06-10 7:04 am
嚇死我,"第一題係'好多年'玩過:",看錯以為 "第一題係'想當年'玩過:"
我仲以為 神貓妙算添,記憶中我好似冇玩過,anyway, 如果我做,可能這樣做:
4^x+6^x=9^x
9^x-6^x-4^x=0
(3/2)^2x-(3/2)^x-1=0
(3/2)^x=(1+√5)/2 或 (1-√5)/2 (不合, 因 (3/2)^x 不可能是負值。)
x log(3/2)=log[(1+√5)/2]
所以 x=[log(1+√5)-log2] / (log3-log2)
2014-06-10 1:36 am
方程式x^3-6x^2+3x+1=0 的三根,試求出a^3+b^3+r^3的值
Sol
a+b+r=6
ab+ar+br=3
a^2+b^2+r^2
=(a+b+r)^2-2(ab+ar+br)
=36-6
=30
x^3-6x^2+3x+1=0
x^3=6x^2-3x-1
a^3=6a^2-3a-1
b^3=6b^2-3b-1
r^3=6r^2-3r-1
a^3+b^3+r^3
=6(a^2+b^2+r^2)-3(a+b+r)-3
=6*30-18-3
=159
2014-06-09 10:42 pm
第一題係好多年玩過:

(4^x) + (6^x) = (9^x)
(4^x)/(9^x) + (6^x)/(9^x) = 1
[(2^x)/(3^x)]² + (2^x)/(3^x) = 1
a² + a = 1 where a = (2/3)^x

Solve a then solve x.

2014-06-09 14:46:10 補充:
第二題:

x³ - 6x² + 3x + 1 = 0 或 x³ = 6x² - 3x - 1 的根為 a, b, r

所以,
a³ = 6a² - 3a - 1
b³ = 6b² - 3b - 1
r³ = 6r² - 3r - 1

a³ + b³ + r³ = 6(a²+b²+r²) - 3(a+b+r) - 3

2014-06-09 14:48:10 補充:
x³ - 6x² + 3x + 1 = 0 的根為 a, b, r

{ a + b + r = 6
{ ab + ar + br = 3
{ abr = -1

考慮
(a + b + r)² = a² + b² + r² + 2(ab + ar + br)
6² = a² + b² + r² + 2(3)
a² + b² + r² = 30

因此,
a³ + b³ + r³ = 6(30) - 3(6) - 3 = 159

2014-06-09 15:12:29 補充:
不不不~

我沒有時間想 2 (b) 和 第 3 題,有待各位繼續~

努力!

╭∧---∧╮
│ .✪‿✪ │
╰/) ⋈ (\╯

2014-06-09 23:54:21 補充:
哈哈哈~
其實係我打錯字~

我係想講係好多年前我都玩過呢題~
仲特登打左個file出黎~

(◕‿◕✿)

2014-06-23 09:18:15 補充:
貼上續帖連結以供大家參考:

(1)
https://hk.knowledge.yahoo.com/question/question?qid=7014060600155

(2)
https://hk.knowledge.yahoo.com/question/question?qid=7014060800065

(3) [本帖]
https://hk.knowledge.yahoo.com/question/question?qid=7014060900051
2014-06-09 10:15 pm
(3)(i) 和提供的條件是一樣的!是打錯了嗎?


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https://hk.answers.yahoo.com/question/index?qid=20140609000051KK00051

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