Innehåll

[redigera] Räknelagar

$a+b=b+a\,\!$	(kommutativa lagen under addition)
$a\cdot b=b\cdot a\,\!$	(kommutativa lagen under multiplikation)
$(a+b)+c=a+(b+c)\,\!$	(associativa lagen under addition)
$(a\cdot b)\cdot c=a\cdot (b\cdot c)\,\!$	(associativa lagen under multiplikation)
$a\cdot (b+c)=a\cdot b+a\cdot c\,\!$	(distributiva lagen)
$a+c=b+c \ \Leftrightarrow \ a=b\,\!$	(annulleringslagen under addition)
$a \cdot c=b \cdot c \ \Leftrightarrow \ a=b \quad om \ c\ne 0\,\!$	(annulleringslagen under multiplikation)

$a\cdot \frac{b}{c}=\frac{a}{1} \cdot \frac{b}{c}=\frac{ab}{c}$	$c\neq 0$
$\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}$	$b\neq 0, d\neq 0$
$\frac{a}{b}\Big/\frac{c}{d}=\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\cdot\frac{d}{c}=\frac{a}{b}\frac{d}{c}=\frac{ad}{bc}$	$b\neq 0, c\neq 0, d\neq 0$
$\frac{a}{b} + \frac{c}{d} = \frac{ad}{bd} + \frac{bc}{bd} = \frac{ad+bc}{bd}$	$b\neq 0, d\neq 0$

$a+(-b)=a-b\,\!$

$a \cdot b=ab\,\!$

$a-(-b)=a+b\,\!$

$a \cdot (-b)=a(-b)=-ab\,\!$

$(-a) \cdot (-b)=(-a)(-b)=ab\,\!$

Låt $a,b,c\in \mathbb{R}$ och $m,n\in \mathbb{Z}$ .

$n! = (1\cdot 2\cdot 3\cdot ...\cdot n) = \prod_{k=1}^n k$	(fakultet)
$(a+b)^n= \sum_{k=0}^{n} {n\choose k}a^{n-k}b^k = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} a^{n-k}b^k$	(binomialteoremet)
$(a_1+a_2+...+a_m)^n= \sum_{k_1+k_2+...+k_m=n}^{} \frac{n!}{{k_1}!{k_2}! ... {k_m}!} a_1^{k_1}a_2^{k_2} ... a_m^{k_m}$	(multinomialteoremet)

$ax+b=0\,\!$	$a \ne 0\,\!$
$x=- \frac {b}{a}\,\!$

Rötterna till andragradsekvationen på formen $x 2 + p x + q = 0$ ges av:

$x_1 = -\frac{p}{2} + \sqrt{\left(\frac{p}{2}\right)^2-q} \quad och \quad x_2 = -\frac{p}{2} - \sqrt{\left(\frac{p}{2}\right)^2-q}$

då gäller

$x_1 + x_2 = -p\,\!$

$x_1 \cdot x_2 = q\,\!$

$a\ och\ b \ge 0,\ c > 0\,\!$

$\sqrt {a} \cdot \sqrt{a}=a\,\!$

$\sqrt {a} \cdot \sqrt{b}= \sqrt {ab}\,\!$

$c \sqrt {a}= \sqrt {c^2 a}\,\!$

$\frac {\sqrt {a}}{\sqrt {c}}= \sqrt{\frac {a}{c}}\,\!$

$\frac {a}{\sqrt {c}}= \frac {a \sqrt {c}}{c}\,\!$

$\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}\,\!$

$\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}\,\!$

$\sqrt[m]{\sqrt[n]{a}}=\sqrt[mn]{a}\,\!$

$a\sqrt[n]{b}=\sqrt[n]{a^n b}\,\!$

$\sqrt[m]{\sqrt[n]{a}}=\sqrt[n]{\sqrt[m]{a}}\,\!$

$\sqrt[nq]{a^{mq}}=\sqrt[n]{a^m}\,\!$

$1^n=1\,\!$
$a^n= \underbrace{a \cdot a \cdot \ldots \cdot a}_{n}$
$a^1 =a\,\!$
$a^0=1\,\!$	$a \ne 0\,\!$
$a^{-n}=\frac{1}{a^n}\,\!$	$a \ne 0\,\!$
$a^{1/2}= \sqrt{a}\,\!$
$a^{m/n}=(a^m)^{1/n}= \sqrt[n]{a^m}\,\!$	$m,\ n\ >0\,\!$
$a^m \cdot a^n=a^{m+n}\,\!$
$\frac{a^m}{a^n}=a^{m-n}\,\!$	$a \ne 0\,\!$
$(ab)^m = a^m\cdot b^m\,\!$
$\left( \frac{a}{b} \right)^m=\frac{a^m}{b^m}\,\!$	$b \ne 0\,\!$
$(a^m)^n=a^{m\cdot n}=(a^n)^m\,\!$
$a^{m/n}=(a^m)^{1/n}= \sqrt[n]{a^m}\,\!$	$m,\ n\ >0\,\!$

$a\ och\ y>0,\ a\ne 1\,\!$

$y=10^x\Leftrightarrow x=\log_{10}\ y=\lg\ y\,\!$

$y=a^x\Leftrightarrow x=\log_{a}\ y\,\!$

$y=e^x\Leftrightarrow x=ln\ y\,\!$

$ln\ y=ln\ 10\cdot lg\ y\approx 2,3026\ lg\ y\,\!$

$lg\ y=lg\ e\cdot ln\ y\approx 0,4343\ ln\ y\,\!$