Question


Answer

Using perfect square and difference of squares formula

(\dfrac{x^2-4x+4}{x^2-1} -\dfrac{x}{x-1} ) \cdotp \dfrac{x-1}{x-2}

=[\dfrac{(x-2)^2}{(x-1)(x+1)}-\dfrac{x}{x-1} ]\cdotp \dfrac{x-1}{x-2}

=\dfrac{x-2}{x+1} -\dfrac{x}{x-2}

=\dfrac{x^2-4x+4-x^2-x}{x^2-x-2}

=\dfrac{4-5x}{x^2-x-2}

Steven Zheng posted 2 weeks ago

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