Does anyone know the answer to this question:?

Question

(1/3) - (2/x-2) + (1/x+1)

(1/3) - (2/x-2) + (1/x+1)

Answer 1

This is not a question, this is an incomplete equation. Where is the right side of the equation, what it is supposed to be equal to?

Answer 2

(1/3) - (2/x-2) + (1/x+1)


= 10/3 - 1/x

Answer 3

(1/3) - (2/x-2) + (1/x+1)

Ok, the first thing I'm doing here is assuming that you want 2/x - 2, not 2/(x-2) where both the xand the -2 are underneath the 2.

First thing you'd do is distribut the negative in the middle, giving:
1/3 - 2/x + 2 +1/x + 1.
Then, -2/x + 1/x = -1/x,
and 1/3 + 2 + 1 = 3and 1/3, or 10/3.
So: -1/x + 10/3.

If your question was where the 2 was over all of x-2, and the 1 was over all of the x+1, then that is not the correct answer.

Answer 4

It depends on whether (2/x-2) is 2/x - 2 or 2/(x-2).

Assuming the latter:
1/3 - 2/(x-2) + 1/(x+1)
(x-2)(x+1) - 3*2(x+1) + 3*(x-2)
x^2 - x - 2 - 6x - 6 + 3x - 6
x^2 - 4x - 14

Answer 5

*Make the last number (1) a fraction: place 1 over 1

(1/3) - (2/x-2) + (1/x + 1/1)

First: find the least common denominator > combine the denominators to get > (3x)(x-2) = 3x^2 - 6x

Sec: multiply the numerators and denominator by the missing terms to get the least common denominator...

*When you multiply a missing term by the denominator, multiply it by the corresponding numerator.

a. [(x)(x-2)(1/3)] - [(3x)(2/x-2)] + [(3)(x-2)(1/x)] + [(3x)(x-2)(1/1)]

b. [(x)(x-2)(1)/(x)(x-2)(3)] - [(3x)(2)/(3x)(x-2)] + [3(x-2)(1)/3(x-2)(x)] + [3x(x-2)(1)/3x(x-2)(1)]

c. (x^2-2x/3x^2-6x) - (6x/3x^2-6x) + (3x-6/3x^2-6x) + (3x^2-6x/3x^2-6)

Third: combine the numerators (keep them in parenthesis) & place them over the least common denominator...

a. [(x^2-2x) - (6x) + (3x-6) + (3x^2-6x)]/(3x^2-6x)

b. (x^2 - 2x - 6x + 3x - 6 + 3x^2 - 6x)/(3x^2-6x)

c. (4x^2 - 11x - 6)/(3x^2-6x)

Answer 6

I think 1/6x

Answer 7

What question did you ask

Answer 8

this is not an equation there is no = sign ( or < or > sign )

it is an expression that has no 'answer'