Hello. Please help me solve this equation, and show what you do, thanks.
x^(2/3) - 5x^(1/3) + 6 = 0
Math Question - Algebra?
2009-09-05 1:35 pm
回答 (5)
2009-09-05 2:06 pm
✔ 最佳答案
(x^1/3 -3)(x^1/3 -2)=0x^1/3=3
x=27 & x^1/3=2 , x=8
God bless you.
2009-09-05 8:57 pm
Y=X^(1/3)
Y^2-5Y+6=0
d=25-24=1
d^1/2=1
Y1=(5-1):2=2
Y2=(5+1):2=3
check:
2^2-5*2+6=4-10+6=0 OK
3^2-5*3+6=9-15+6=0 OK
so
X1=2^3=8
X2=3^3=27
Y^2-5Y+6=0
d=25-24=1
d^1/2=1
Y1=(5-1):2=2
Y2=(5+1):2=3
check:
2^2-5*2+6=4-10+6=0 OK
3^2-5*3+6=9-15+6=0 OK
so
X1=2^3=8
X2=3^3=27
2009-09-05 8:40 pm
upper answer is wrong..i'm sorry..
Y=X^(1/3)
then the equation becomes..
Y^2-5Y+6=0
so Y=2,3
so X=8,27
Y=X^(1/3)
then the equation becomes..
Y^2-5Y+6=0
so Y=2,3
so X=8,27
2009-09-05 11:52 pm
x^(2/3) - 5x^(1/3) + 6 = 0
x^(2/3) - 2x^(1/3) - 3x^(1/3) + 6 = 0
[x^(2/3) - 2x^(1/3)] - [3x^(1/3) - 6] = 0
x^(1/3)[x^(1/3) - 2] - 3[x^(1/3) - 2] = 0
[x^(1/3) - 2][x^(1/3) - 3] = 0
x^(1/3) - 2 = 0
x^(1/3) = 2
x = 2^3
x = 8
x^(1/3) - 3 = 0
x^(1/3) = 3
x = 3^3
x = 27
â´ x = 8, 27
x^(2/3) - 2x^(1/3) - 3x^(1/3) + 6 = 0
[x^(2/3) - 2x^(1/3)] - [3x^(1/3) - 6] = 0
x^(1/3)[x^(1/3) - 2] - 3[x^(1/3) - 2] = 0
[x^(1/3) - 2][x^(1/3) - 3] = 0
x^(1/3) - 2 = 0
x^(1/3) = 2
x = 2^3
x = 8
x^(1/3) - 3 = 0
x^(1/3) = 3
x = 3^3
x = 27
â´ x = 8, 27
2009-09-05 8:40 pm
2x/3-5x/3+6=0
(2x-5x+2)/3=0
2-3x=0
x=2/3
this is wrong.
(2x-5x+2)/3=0
2-3x=0
x=2/3
this is wrong.
收錄日期: 2021-05-01 12:45:11
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https://hk.answers.yahoo.com/question/index?qid=20090905053515AAp6iN9