Álgebra básica

ASOCIATIVA:    $a(bc) = (ab)c$
CONMUTATIVA:  $a+b=b+c$ y $ab = bc$
DISTRIBUTIVA: $a(b+c) = ab + ac$

$$ab+ac=a (b +c)$$
$$\begin{eqnarray}a(\frac{b}{c})&=\frac{ab}{c}\\ \frac{(\frac{a}{b})}{c} &= \frac{a}{bc}\\ \frac{a}{(\frac{b}{c})} &=\frac{ac}{b}\\ \frac{(\frac{a}{b})}{(\frac{c}{d})} &= \frac{ad}{bc}\end{eqnarray}$$	$$\begin{eqnarray}\frac{a}{b}+\frac{c}{d} &= \frac{ad+bc}{bd}\\ \frac{a}{b}-\frac{c}{d} &= \frac{ad-bc}{bd}\\ \frac{a-b}{c-d}&=\frac{b-a}{d-c}\\  \frac{a+b}{c}&=\frac{a}{c}+\frac{b}{c}\end{eqnarray}$$
$$\frac{ab+ac}{a}=b+c, a\neq 0$$

$$\begin{eqnarray}\sqrt[n]{a}&=a^{\frac{1}{n}}\\ \sqrt[m]{\sqrt[n]{a}}&=\sqrt[nm]{a}\end{eqnarray}$$	$$\begin{eqnarray}\sqrt[n]{ab}&=\sqrt[n]{a}\sqrt[n]{b}\\ \sqrt[n]{\frac{a}{b}}&=\frac{\sqrt[n]{a}}{\sqrt[n]{b}} \end{eqnarray}$$
$$\begin{eqnarray}\sqrt[n]{a^n} = a&, n=2k+1\\ \sqrt[n]{a^n} = |a|&, n=2k\end{eqnarray}$$

$$|a| = \left\{\begin{array}{rl}  a & \text{si } a \geq 0\\   -a & \text{si } a > 0 \end{array} \right.$$	$$\begin{eqnarray}|a|&=|-a| \\ |ab|&=|a||b|\end{eqnarray}$$
DESIGUALDAD DEL TRIÁNGULO $$|a+b| \leq |a|+ |b|$$	$$\begin{eqnarray}|a| &\geq 0\\ \big|\frac{a}{b}\big| &=\frac{|a|}{|b|}\end{eqnarray}$$

$$\begin{eqnarray}a^{n}a^{m}&=a^{n+m}\\ (ab)^{n}&=a^{n}b^{n}\\ a^{-n}&=\frac{1}{a^n}\\ a^{0}=1&, a \neq 0\end{eqnarray}$$	$$\begin{eqnarray}{(a^{n})}^{m}&=a^{nm}\\ (\frac{a}{b})^n&=\frac{a^n}{b^n}\\ \frac{1}{a^{-n}}&=a^n\\ (\frac{a}{b})^n&=\frac{a^n}{b^n}\end{eqnarray}$$
$$\frac{a^{n}}{a^{m}}=a^{n-m}=\frac{1}{a^{m-n}}$$ $$(a)^{\frac{n}{m}}=(a^{\frac{1}{m}})^n=(a^n)^{\frac{1}{m}}$$ $$(\frac{a}{b})^{-n} = (\frac{b}{a})^n = \frac{b^n}{a^n}$$

 Para resolver ecuaciones cuadráticas

$$ax^2+bx+c=0$$ utilizamos la fórmula general $$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$

$$\begin{eqnarray}x^2-a^2&= (x+a)(x-a)\\ x^2+(a+b)x + ab &= (x+a)(x+b)\\ \end{eqnarray}$$
$$\begin{eqnarray}x^2 +2ax + a^2 &= (x+a)^2 \\ x^2 -2ax +a^2 &= (x-a)^2\\ x^3 + 3ax^2 + 3a^2x + a^3 &= (x+a)^3\\  x^3-3ax^2+3a^2x-a^3 &= (x-a)^3 \end{eqnarray}$$
$$\begin{eqnarray}x^3+a^3 &= (x+a)(x^2-ax+a^2)\\ x^3-a^3 &= (x-a)(x^2+ax+a^2) \end{eqnarray}$$