How do you combine 1/x + 2/x^2 - 3/(x+x^2)?

Nov 3, 2015

$\textcolor{b l u e}{\frac{x \left(x + {x}^{2}\right) + 2 \left(x + {x}^{2}\right) + 3 {x}^{2}}{{x}^{2} \left(x + {x}^{2}\right)}}$
I will let you work this out.
Detailed explanation will help in solving others.

Explanation:

You need common a denominator. That is something that each of the separate denominators will divide into without having a remainder (fractional component)

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Consider the denominators od $x , {x}^{2} \text{ and } x + {x}^{2}$

$x + {x}^{2} \text{ }$will divide into itself

${x}^{2}$ will not divide into $x + {x}^{2}$ without a remainder. So we build something it will divide into. That is ${x}^{2} \left(x + {x}^{2}\right)$

$x$ will divide into the ${x}^{2}$ part of ${x}^{2} \left(x + {x}^{2}\right)$

$\textcolor{red}{\text{Using " x^2(x+x^2) " as the common denominator we have:}}$

For $\frac{1}{x}$ we need to make this look like $\frac{\text{something}}{{x}^{2} \left(x + {x}^{2}\right)}$ but without changing its true value. This is done by multiplying by 1 but in the form of $\frac{x \left(x + {x}^{2}\right)}{x \left(x + {x}^{2}\right)}$

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For $\textcolor{b l u e}{\frac{1}{x}}$ we have $\frac{1}{x} \times \frac{x \left(x + {x}^{2}\right)}{x \left(x + {x}^{2}\right)} = \textcolor{b l u e}{\frac{x \left(x + {x}^{2}\right)}{{x}^{2} \left(x + {x}^{2}\right)}}$

For $\textcolor{b l u e}{\frac{2}{x} ^ 2}$ we have $\frac{2}{x} ^ 2 \times \frac{x + {x}^{2}}{x + {x}^{2}} = \textcolor{b l u e}{\frac{2 \left(x + {x}^{2}\right)}{{x}^{2} \left(x + {x}^{2}\right)}}$

For $\textcolor{b l u e}{\frac{3}{x + {x}^{2}}}$ we have 3/(x+x^2) times x^2/x^2 =color(blue)((3x^2)/(x^2(x+x^2))

Putting it all together

$\textcolor{b l u e}{\frac{x \left(x + {x}^{2}\right) + 2 \left(x + {x}^{2}\right) + 3 {x}^{2}}{{x}^{2} \left(x + {x}^{2}\right)}}$