請問:x^4-3x^3-2x^2-3x+1=0，求x=??
回答 (5)
2015-11-29 8:03 am
✔ 最佳答案
令原式=(x^2+ax+1)(x^2+bx+1)=0==>x^4+(a+b)x^3+(2+ab)x^2+(a+b)x+1=x^4-3x^3-2x^2-3x+1=0
比較係數知:a+b=-3,2+ab=-2
-==>a=-4,b=1或a=1,b=-4
x^4-3x^2-2x^2-3x+1=0
==>(x^2-4x+1)(x^2+x+1)=0
==>x=2+-根號3或 x=(-1+-根號3 i )/2
令原式=(x^2+ax-1)(x^2+bx-1)=0
同上面作法不合
2015-12-25 1:40 pm
將x=0代入:0-0-0-0+1=1，故不是0
=>x^2-3x-2-3/x+1/x^2=0(兩邊同除x^2)
=>[x^2+2-2+(1/x^2)]+(-3x-3/x)-2=0(各自做分組)
=>{x^2+2*x*(1/x)+(1/x)^2]-2-3[x+(1/x)]-2(試著將內部分解)
=>[x+(1/x)]^2-3[x+(1/x)]-4
設x+(1/x)=A
=>A^2-3A-4=0(代入A)
=>(A+1)(A-4)=0(十字交乘)
x+(1/x)+1=0 x+(1/x)-4=0(帶入A)
x不等於0=>x^2+x+1=0 x^2-4x+1=0
x=(-1± √3i)或2± √3
=>x^2-3x-2-3/x+1/x^2=0(兩邊同除x^2)
=>[x^2+2-2+(1/x^2)]+(-3x-3/x)-2=0(各自做分組)
=>{x^2+2*x*(1/x)+(1/x)^2]-2-3[x+(1/x)]-2(試著將內部分解)
=>[x+(1/x)]^2-3[x+(1/x)]-4
設x+(1/x)=A
=>A^2-3A-4=0(代入A)
=>(A+1)(A-4)=0(十字交乘)
x+(1/x)+1=0 x+(1/x)-4=0(帶入A)
x不等於0=>x^2+x+1=0 x^2-4x+1=0
x=(-1± √3i)或2± √3
參考: 自己
2015-12-09 5:53 am
x^4-3x^3-2x^2-3x+1=0
x=3.732050807568877
or
x=0.2679491924311228
or
x=-0.5+0.8660254037844386i
or
x=-0.5-0.8660254037844386i
x=3.732050807568877
or
x=0.2679491924311228
or
x=-0.5+0.8660254037844386i
or
x=-0.5-0.8660254037844386i
2015-11-30 5:59 am
x=0,3.7473465
工程計算機
用手算的話最後變成 x[x(x-2)(x-1)-3]=-1, 會變很麻煩
工程計算機
用手算的話最後變成 x[x(x-2)(x-1)-3]=-1, 會變很麻煩
2015-11-29 10:08 am
x^4-3x^3-2x^2-3x+1=0
((x-4)x+1)(x^2+x+1)=0
(x^2-4x+1)(x^2+x+1)=0
x^4+1 = x(3x^2+2x+3)
x = 2 ± √3
((x-4)x+1)(x^2+x+1)=0
(x^2-4x+1)(x^2+x+1)=0
x^4+1 = x(3x^2+2x+3)
x = 2 ± √3
收錄日期: 2021-04-18 14:05:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151128233344AAEB9u1