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# ((x-2)/(2x)+(1)/(x+2))/((3)/(2)-(6)/(x^(2)+2x))

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((x-2)/(2x)+(1)/(x+2))/((3)/(2)-(6)/(x^(2)+2x))

How would I solve this?

Please give step by step directions on how to solve

Guest Oct 17, 2017
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#1
+18715
+2

How would I solve this?

((x-2)/(2x)+(1)/(x+2))/((3)/(2)-(6)/(x^(2)+2x))

$$\begin{array}{|rcll|} \hline && \mathbf{\dfrac{ \dfrac{x-2}{2x} + \dfrac{1}{x+2} } { \dfrac{3}{2}- \dfrac{6} {x^{2}+2x} } } \quad & | \quad x^{2}+2x = x(x+2) \\\\ &=& \dfrac{ \dfrac{x-2}{2x} + \dfrac{1}{x+2} } { \dfrac{3}{2}- \dfrac{6} {x(x+2)} } \\\\ &=& \dfrac{ \dfrac{(x-2)(x+2) + 1\cdot 2x}{2x(x+2)} } { \dfrac{3x(x+2)-2\cdot 6}{2x(x+2)} } \\\\ &=& \dfrac{\left[~(x-2)(x+2) + 1\cdot 2x ~\right]}{2x(x+2)} \cdot \dfrac{2x(x+2)} {\left[~ 3x(x+2)-2\cdot 6 ~\right] } \\\\ &=& \dfrac{(x-2)(x+2) + 1\cdot 2x} {3x(x+2)-2\cdot 6} \\\\ &=& \dfrac{(x-2)(x+2) + 2x} {3x(x+2)-12} \\\\ &=& \dfrac{x^2-4 + 2x } {3x^2+6x-12} \\\\ &=& \dfrac{1}{3} \cdot \dfrac{\left(x^2+2x-4\right)} {\left(x^2+2x-4\right)} \\\\ &\mathbf{=}& \mathbf{ \dfrac{1}{3}} \\ \hline \end{array}$$

heureka  Oct 17, 2017
#2
0

i dont understand the last two steps can u pls explain

Guest Oct 18, 2017
#3
+18715
+3

i dont understand the last two steps can u pls explain

$$\begin{array}{|rcll|} \hline && \dfrac{{\color{red}(x-2)(x+2)} + 2x} {3x(x+2)-12} \quad & | \quad {\color{red}(x-2)(x+2)} = x^2 + 2x-2x- 2\cdot 2 = x^2 -4 \\\\ &=& \dfrac{x^2-4 + 2x } {3x^2+6x-12} \\\\ &=& \dfrac{x^2-4 + 2x } {{\color{red}3}x^2+2\cdot {\color{red}3}\cdot x-({\color{red}3}\cdot 4)} \\\\ &=& \dfrac{x^2-4 + 2x } { {\color{red}3}\cdot ( x^2+2\cdot x-( 4))} \\\\ &=& \dfrac{x^2-4 + 2x } { {\color{red}3}\cdot ( x^2-4+2x)} \\\\ &=& \dfrac{(x^2-4 + 2x) } { {\color{red}3}\cdot ( x^2-4+2x)} \quad &| \quad \dfrac{(x^2-4 + 2x) } { ( x^2-4+2x)} = 1 \\\\ &=& \dfrac{1} { {\color{red}3}} \cdot 1 \\\\ &=& \dfrac{1} { {\color{red}3}} \\\\ \hline \end{array}$$

heureka  Oct 18, 2017

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