x-4/5=1/x
x does not equal O
help!!
and can you explain it to me?
solve for the positive value of x?
2009-04-17 3:34 pm
回答 (10)
2009-04-17 3:40 pm
✔ 最佳答案
Mutiply both sides by 5x,x(x-4) = 5
x^2-4x-5 = (x-5)(x+1) = 0
x = 5, the positive root.
2009-04-17 3:51 pm
(x - 4)/5 = 1/x
x² - 4x = 5
x² - 2x = 5 + (- 2)²
x² - 2x = 5 + 4
(x - 2)² = 9
x - 2 = 3
x = 3 + 2, x = 5
x = - 3 + 2, x = - 1
Answer: x = 5, - 1
Proof (x = 5):
(5 - 4)/5 = 1/5
1/5 = 1/5
Proof (x = - 1):
(- 1 - 4)/5 = 1/ - 1
- 5/5 = - 1
- 1 = - 1
x² - 4x = 5
x² - 2x = 5 + (- 2)²
x² - 2x = 5 + 4
(x - 2)² = 9
x - 2 = 3
x = 3 + 2, x = 5
x = - 3 + 2, x = - 1
Answer: x = 5, - 1
Proof (x = 5):
(5 - 4)/5 = 1/5
1/5 = 1/5
Proof (x = - 1):
(- 1 - 4)/5 = 1/ - 1
- 5/5 = - 1
- 1 = - 1
2009-04-18 8:34 pm
These questions are always suspect due to lack of brackets.
Do you actually mean :-
(x - 4) / 5 = 1 / x__________????????
x² - 4x = 5
x² - 4x - 5 = 0
(x - 5)(x + 1) = 0
x = 5
PS
Should not have to GUESS the question !
Do you actually mean :-
(x - 4) / 5 = 1 / x__________????????
x² - 4x = 5
x² - 4x - 5 = 0
(x - 5)(x + 1) = 0
x = 5
PS
Should not have to GUESS the question !
2009-04-18 11:13 am
(x - 4)/5 = 1/x
x - 4 = 5/x
x(x - 4) = 5
x^2 - 4x = 5
x^2 - 4x - 5 = 0
x^2 + x - 5x - 5 = 0
(x^2 + x) - (5x + 5) = 0
x(x + 1) - 5(x + 1) = 0
(x + 1)(x - 5) = 0
x + 1 = 0
x = -1
x - 5 = 0
x = 5
∴ x = -1 , 5
x - 4 = 5/x
x(x - 4) = 5
x^2 - 4x = 5
x^2 - 4x - 5 = 0
x^2 + x - 5x - 5 = 0
(x^2 + x) - (5x + 5) = 0
x(x + 1) - 5(x + 1) = 0
(x + 1)(x - 5) = 0
x + 1 = 0
x = -1
x - 5 = 0
x = 5
∴ x = -1 , 5
2009-04-17 3:55 pm
x-4/5 = 1/x
LCM is 5x
x(x-4) = 5
distributive property
x^2 - 4x - 5 = 0
factor
(x-5)(x+1) = 0
x= -1 and 5
That's it Maths rock
LCM is 5x
x(x-4) = 5
distributive property
x^2 - 4x - 5 = 0
factor
(x-5)(x+1) = 0
x= -1 and 5
That's it Maths rock
2009-04-17 3:45 pm
this equation has no solution:
http://www.mathway.com/answer.aspx?p=trig?p=x-4SMB105SMB011SMB10xSMB15?p=1?p=?p=?p=?p=?p=?p=?p=0?p=?p=
http://www.mathway.com/answer.aspx?p=trig?p=x-4SMB105SMB011SMB10xSMB15?p=1?p=?p=?p=?p=?p=?p=?p=0?p=?p=
2009-04-17 3:42 pm
x-4/5=1/x
multiply the whole equation by "x"
x^2 - (4/5)*x = 1
transpose "1"
x^2 - (4/5)*x -1 = 0
solving equation:
2/5+1/5*29^(1/2)
2/5-1/5*29^(1/2)
positive value of x:
2/5+1/5*29^(1/2) = 1.4770
THERE ARE DIFFERENT BETWEEN
x-4/5=1/x
AND
(x-4)/5=1/x
multiply the whole equation by "x"
x^2 - (4/5)*x = 1
transpose "1"
x^2 - (4/5)*x -1 = 0
solving equation:
2/5+1/5*29^(1/2)
2/5-1/5*29^(1/2)
positive value of x:
2/5+1/5*29^(1/2) = 1.4770
THERE ARE DIFFERENT BETWEEN
x-4/5=1/x
AND
(x-4)/5=1/x
2009-04-17 3:41 pm
First step:
Multiply everything by x
= X^2 - (4x/5) = 1
Second Step:
Transpose to a quadratic
= X^2 - (4x/5) - 1 = 0
Third Step:
Multiply by 5
= 5x^2 - 4x - 5 = 0
Fourth step:
Solve for x
Multiply everything by x
= X^2 - (4x/5) = 1
Second Step:
Transpose to a quadratic
= X^2 - (4x/5) - 1 = 0
Third Step:
Multiply by 5
= 5x^2 - 4x - 5 = 0
Fourth step:
Solve for x
2009-04-17 3:40 pm
We want to get everything on the left side of the equation...
(x - 4) / 5 = 1 / x
Multiply throughout by 5
(x - 4) = 5 / x
Multiply throughout by x
x (x - 4) = 5
Expand the brackets
x^2 - 4x = 5
Bring the 5 to the left side
x^2 - 4x -5 = 0
Factor it
(x + 1)(x - 5) = 0
x = -1 or x = 5
(x - 4) / 5 = 1 / x
Multiply throughout by 5
(x - 4) = 5 / x
Multiply throughout by x
x (x - 4) = 5
Expand the brackets
x^2 - 4x = 5
Bring the 5 to the left side
x^2 - 4x -5 = 0
Factor it
(x + 1)(x - 5) = 0
x = -1 or x = 5
2009-04-17 3:40 pm
Multiply by x
Then line two is Xsquared -0.8X =1
Then third line is Xsquared -0.8X - 1 = 0
You've got a formual for that?
All I did now, was to manipulate it into the style the formula can solve, you noticed this?
Then line two is Xsquared -0.8X =1
Then third line is Xsquared -0.8X - 1 = 0
You've got a formual for that?
All I did now, was to manipulate it into the style the formula can solve, you noticed this?
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