設a+b=3，ab=1，求a^2+b^2和a^3+b^3還有a-b 設x-1/x=4，則x^3-1/x^3=?
回答 (2)
2015-09-06 3:39 pm
1.
a^2+b^2=(a+b)^2-2ab=3^2-2*1=7
a^3+b^3=(a+b)(a^2-ab+b^2)=3*(7-1)=18
(a-b)^2=(a+b)^2-4ab=3^2-4*1=5
==>a-b=+-根號5
2.
x-(1/x)=4
==>(x-(1/x))^2=16
==>x^2+(1/x^2)=18
x^3-(1/x^3)=(x-(1/x))(x^2+1+(1/x^2))=4*(18+1)=76
a^2+b^2=(a+b)^2-2ab=3^2-2*1=7
a^3+b^3=(a+b)(a^2-ab+b^2)=3*(7-1)=18
(a-b)^2=(a+b)^2-4ab=3^2-4*1=5
==>a-b=+-根號5
2.
x-(1/x)=4
==>(x-(1/x))^2=16
==>x^2+(1/x^2)=18
x^3-(1/x^3)=(x-(1/x))(x^2+1+(1/x^2))=4*(18+1)=76
2015-12-04 3:18 pm
a+b=3
a^2+2ab+b^2=9
a^2+b^2=7
a+b=3
(a+b)^3 =27
a^3+3a^2 b+3ab^2 +b^3=27
a^3+b^3 +3ab(a+b)=27
a^3+b^3+3(3)=27
a^3+b^3=18
(x-1)/x=4
x-1=4x
-1=3x
x=-1/3
(x^3-1)/x^3
=[(1/3)^3-1]/(1/3)^3
=[(1/27)-1]/(1/27)
=1-27
=-26
a^2+2ab+b^2=9
a^2+b^2=7
a+b=3
(a+b)^3 =27
a^3+3a^2 b+3ab^2 +b^3=27
a^3+b^3 +3ab(a+b)=27
a^3+b^3+3(3)=27
a^3+b^3=18
(x-1)/x=4
x-1=4x
-1=3x
x=-1/3
(x^3-1)/x^3
=[(1/3)^3-1]/(1/3)^3
=[(1/27)-1]/(1/27)
=1-27
=-26
收錄日期: 2021-04-18 00:10:07
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150906072832AA66zli