若x=∛[a+√(a²+b³)] - ∛[√(a²+b³)-a], 試以a表示x³+3bx?
回答 (2)
2015-11-25 6:03 am
✔ 最佳答案
Solp=[a+√(a^2+b^3)]^(1/3)
q=[√(a^2+b^3)－a]^(1/3)
p^3=a+√(a^2+b^3)
q^3=√(a^2+b^3)－a
p^3－q^3=2a
pq=[√(a^2+b^3)^2－a^2]^(1/3)=b
x=p－q
x^3=p^3－q^3－3pq(p－q)
=2a－3bx
x^3+3bx=2a
2015-12-14 12:10 pm
p=[a+√(a^2+b^3)]^(1/3)
q=[√(a^2+b^3)－a]^(1/3)
p^3=a+√(a^2+b^3)
q^3=√(a^2+b^3)－a
p^3－q^3=2a
pq=[√(a^2+b^3)^2－a^2]^(1/3)=b
x=p－q
x^3=p^3－q^3－3pq(p－q)
=2a－3bx
x^3+3bx=2a
a=(x^3+3bx)/2
q=[√(a^2+b^3)－a]^(1/3)
p^3=a+√(a^2+b^3)
q^3=√(a^2+b^3)－a
p^3－q^3=2a
pq=[√(a^2+b^3)^2－a^2]^(1/3)=b
x=p－q
x^3=p^3－q^3－3pq(p－q)
=2a－3bx
x^3+3bx=2a
a=(x^3+3bx)/2
收錄日期: 2021-04-18 14:08:07
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151124184513AAg1vQa