Question 153924
{{{(1)/(x^2-2x)-5/(x^2-4x+4)}}} Start with the given expression



{{{(1)/(x(x-2))-5/(x^2-4x+4)}}} Factor {{{x^2-2x}}} to get {{{x(x-2)}}}



{{{(1)/(x(x-2))-5/((x-2)(x-2))}}} Factor {{{x^2-4x+4}}} to get {{{(x-2)(x-2)}}}



Notice how the LCD is {{{x(x-2)(x-2)}}}



{{{((x-2)/(x-2))((1)/(x(x-2)))-5/((x-2)(x-2))}}} Multiply the first fraction by {{{(x-2)/(x-2)}}}



{{{(x-2)/(x(x-2)(x-2))-5/((x-2)(x-2))}}} Combine and multiply the fractions



{{{(x-2)/(x(x-2)(x-2))-(x/x)(5/((x-2)(x-2)))}}} Multiply the second fraction by {{{(x)/(x)}}}



{{{(x-2)/(x(x-2)(x-2))-(5x)/(x(x-2)(x-2))}}} Combine and multiply the fractions




Since the denominators are now equal, this means that we can subtract the fractions. To do this, simply subtract the numerators and place that over the common denominator.



{{{(x-2-5x)/(x(x-2)(x-2))}}}



{{{(-6x-2)/(x(x-2)(x-2))}}} Combine like terms.




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Answer:


So {{{(1)/(x^2-2x)-(5)/(x^2-4x+4)}}} subtracts and simplifies to {{{(-6x-2)/(x(x-2)(x-2))}}}