Question 1189175
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Let's factor the denominators and do a bit of tidying up.


{{{x/(2x^2-7x+3) - (x-4) /((2x-1)(3-x)) - 7 / (x^2-x-6)}}}


{{{x/((2x-1)(x-3)) - (x-4) /((2x-1)(-(x-3))) - 7 / ((x-3)(x+2))}}}


{{{x/((2x-1)(x-3)) - (x-4) /(-(2x-1)(x-3)) - 7 / ((x-3)(x+2))}}}


{{{x/((2x-1)(x-3)) + (x-4) /((2x-1)(x-3)) - 7 / ((x-3)(x+2))}}}


{{{(x + (x-4)) /((2x-1)(x-3)) - 7 / ((x-3)(x+2))}}}


{{{(2x-4) /((2x-1)(x-3)) - 7 / ((x-3)(x+2))}}}


When focusing on the denominators, the unique factors are: (2x-1), (x-3), (x+2)
The LCD is therefore (2x-1)(x-3)(x+2)
Let's get each denominator to be equal to the LCD. That way we can add the fractions.


{{{(2x-4) /((2x-1)(x-3)) - 7 / ((x-3)(x+2))}}}


{{{((2x-4) /((2x-1)(x-3)))*((x+2)/(x+2)) - (7 / ((x-3)(x+2)))*((2x-1)/(2x-1))}}}


{{{((2x-4)(x+2)) /((2x-1)(x-3)(x+2)) - (7(2x-1)) / ((2x-1)(x-3)(x+2))}}}


{{{(2x^2-8) /((2x-1)(x-3)(x+2)) - (14x-7) / ((2x-1)(x-3)(x+2))}}}


{{{(2x^2-8 - (14x-7) )/ ((2x-1)(x-3)(x+2))}}}


{{{(2x^2-8 - 14x+7 )/ ((2x-1)(x-3)(x+2))}}}


{{{(2x^2 - 14x-1 )/ ((2x-1)(x-3)(x+2))}}}


{{{(2x^2 - 14x-1 )/ ((2x-1)(x^2-x-6))}}}


{{{(2x^2 - 14x-1 )/ (2x(x^2-x-6)-1(x^2-x-6))}}}


{{{(2x^2 - 14x-1 )/ (2x^3-2x^2-12x-x^2+x+6)}}}


{{{(2x^2 - 14x-1 )/ (2x^3-3x^2-11x+6)}}}


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Answer: The original expression simplifies fully to {{{(2x^2 - 14x-1 )/ (2x^3-3x^2-11x+6)}}}
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