Thread: Operations on Real Numbers and Rational Equations

1. Operations on Real Numbers and Rational Equations

$\displaystyle [(x^2n + x^n)/(2x - 2)] / [(4x^n + 4)/(x^n+1 - x^n)]$

NOTE: $\displaystyle (x^2n + x^n)$ : The power isn't 2, it's 2n.
$\displaystyle (x^n+1 - x^n)$ : The power isn't n, it's n+1.

Sorry, I don't know how to get it to work properly.

$\displaystyle [(x-1)/(x+1) - (x+1)/(x-1)] / [(x-1)/(x+1) + (x+1)/(x-1)]$

Last two problems.

2. [(x^2n + x^n)/(2x - 2)] / [(4x^n + 4)/(x^n+1 - x^n)]
$\displaystyle \frac {\frac {x^{2n}+x^n}{2x-2}}{\frac {4x^n+4}{x^{n+1}-x^n}}$

to get more than one thingy in a superscript, surround them with curly brackets eg x^{2n}.

Just factorise all of the brackets then cancel. Remember that $\displaystyle x^{2n} = x^n*x^n$ and $\displaystyle \frac {a/b}{c/d} = \frac {ad}{bc}$

3. $\displaystyle \frac{{\frac{{x^{2n} + x^n }} {{2x - 2}}}} {{\frac{{4x^n + 4}} {{x^{n + 1} - x^n }}}} = \frac{{x^{2n} + x^n }} {{2x - 2}} \cdot \frac{{x^{n + 1} - x^n }} {{4x^n + 4}} = \frac{{x^n \left( {x^n + 1} \right)}} {{2\left( {x - 1} \right)}} \cdot \frac{{x^n \left( {x - 1} \right)}} {{4\left( {x^n + 1} \right)}} = \frac{{x^n x^n }} {{\left( 2 \right)\left( 4 \right)}} = \frac{{x^{2n} }} {8}$

$\displaystyle \boxed{ {\text{Let }}a = x - 1\text{; } {\text{ }}b = x + 1 }$:

$\displaystyle \frac{{\frac{a} {b} - \frac{b} {a}}} {{\frac{a} {b} + \frac{b} {a}}} = \frac{{\frac{{a^2 - b^2 }} {{ab}}}} {{\frac{{a^2 + b^2 }} {{ab}}}}$$\displaystyle = \frac{{a^2 - b^2 }} {{ab}} \cdot \frac{{ab}} {{a^2 + b^2 }} = \frac{{a^2 - b^2 }} {{a^2 + b^2 }} = \frac{{\left( {a + b} \right)\left( {a - b} \right)}} {{a^2 + b^2 }} \displaystyle \frac{{\left( {a + b} \right)\left( {a - b} \right)}} {{a^2 + b^2 }} =$$\displaystyle \frac{{\left( {\left( {x - 1} \right) + \left( {x + 1} \right)} \right)\left( {\left( {x - 1} \right) - \left( {x + 1} \right)} \right)}} {{\left( {x - 1} \right)^2 + \left( {x + 1} \right)^2 }} = $$\displaystyle \frac{{\left( {2x} \right)\left( { - 2} \right)}} {{\left( {x^2 - 2x + 1} \right) + \left( {x^2 + 2x + 1} \right)}} =$$\displaystyle \frac{{ - 4x}} {{2x^2 + 2}} = \frac{{ - 2x}} {{x^2 + 1}}$