1. [ (2x/x+3) - (x+2/x) / (x^2 + 3x + 2/x^3+2x^2-3x) ] - (6-9x/ x+2)
2x over x+3 minus x+2 over x etc that's how it goes
2. [(x - 3overx-2 ) ( 1over x+1 - 1)] - 1overx-2 - 1over(x-2)(x-2)
3. [ 1 over x^2-3x+2 + x^2+2xoverx^2-4 ] / x^3+ 1over2x^2-2x-4
tnx in advance
I try to guess what you mean:Originally Posted by eepyej
Factorize the terms in the denominator and add the fractions of the numerator:
Expand the bracket in the numerator and collect like terms. Then multiply by the reciprocal of the denominator:
Factorize the numerator of the first factor and cancel out like terms:
\rlap{\color{red}====}(x+1)}{\rlap{=====}x(x+3)} \cdot \dfrac{\rlap{=====}x(x+3)(x-1)}{(x+2)\rlap{\color{red}====}(x+1)} - \dfrac{6-9x}{x+2})
Obviously my guess seems to be correct.
By the way: To prevent any wild guesses you should use parantheses. Your problem should have been written like:
[ (2x/x+3) - (x+2/x) / (x^2 + 3x + 2/x^3+2x^2-3x) ] - (6-9x/ x+2) = (2x/(x+3))-((x+2)/x /((x^2+3x+2)/(x^3+2x^2-3x))) )-(6-9x)/(x+2)
I assume that there is a typo. I'll do the problem as written above:
Simplify the term in the second paranthesis and the last 2 summands:
\left( -\frac{x}{x+1} \right)-\left(\frac{x+2}{(x-2)(x+1)}\right))
(x+1)})
And now I'm doing the original version:
Simplify the term in the second paranthesis and the last 2 summands:
The common denominator is
. Therefore the fractions become:
Expand the brackets and collect like terms:
-(x^2-1)}{(x+1)(x-2)^2} = -\frac{x^3-4x^2+6x-1}{(x+1)(x-2)^2})
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