Question

\dfrac{3x+6}{x^2+4x+4}÷\dfrac{x-2}{x+2} -\dfrac{1}{x-2} , in which

x=-6

Steven Zheng Steven Zheng posted 2 weeks ago


Answer

First simplify the expression

\dfrac{3x+6}{x^2+4x+4}÷\dfrac{x-2}{x+2} -\dfrac{1}{x-2}

=\dfrac{3(x+2)}{(x+2)^2}\cdotp \dfrac{x+2}{x-2} -\dfrac{1}{x-2}

=\dfrac{3}{x-2} -\dfrac{1}{x-2} =\dfrac{2}{x-2}=-\dfrac{1}{4}

Steven Zheng Steven Zheng posted 2 weeks ago

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