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# Formelsamling/Matematik/Algebra

## Räknelagar

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

### Bråkregler

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

### Parentesregler

 ${\displaystyle a+(-b)=a-b\,\!}$ ${\displaystyle a\cdot b=ab\,\!}$ ${\displaystyle a-(-b)=a+b\,\!}$ ${\displaystyle a\cdot (-b)=a(-b)=-ab\,\!}$ ${\displaystyle (-a)\cdot (-b)=(-a)(-b)=ab\,\!}$

## Algebra

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

 ${\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}\,\!}$ (första kvadreringsregeln) ${\displaystyle (a-b)^{2}=a^{2}-2ab+b^{2}\,\!}$ (andra kvadreringsregeln) ${\displaystyle (a+b)(a-b)=a^{2}-b^{2}\,\!}$ (konjugatregeln) ${\displaystyle (a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}\,\!}$ ${\displaystyle (a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}\,\!}$ ${\displaystyle a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})\,\!}$ ${\displaystyle a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})\,\!}$

 ${\displaystyle n!=(1\cdot 2\cdot 3\cdot ...\cdot n)=\prod _{k=1}^{n}k}$ (fakultet) ${\displaystyle (a+b)^{n}=\sum _{k=0}^{n}{n \choose k}a^{n-k}b^{k}=\sum _{k=0}^{n}{\frac {n!}{k!(n-k)!}}a^{n-k}b^{k}}$ (binomialteoremet) ${\displaystyle (a_{1}+a_{2}+...+a_{m})^{n}=\sum _{k_{1}+k_{2}+...+k_{m}=n}^{}{\frac {n!}{{k_{1}}!{k_{2}}!...{k_{m}}!}}a_{1}^{k_{1}}a_{2}^{k_{2}}...a_{m}^{k_{m}}}$ (multinomialteoremet)

${\displaystyle x^{2}+px=x^{2}+px+({\frac {p}{2}})^{2}-({\frac {p}{2}})^{2}=(x+{\frac {p}{2}})^{2}-({\frac {p}{2}})^{2}}$

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

Rötterna till andragradsekvationen på formen ${\displaystyle x^{2}+px+q=0}$ ges av:

${\displaystyle x_{1}=-{\frac {p}{2}}+{\sqrt {\left({\frac {p}{2}}\right)^{2}-q}}\quad och\quad x_{2}=-{\frac {p}{2}}-{\sqrt {\left({\frac {p}{2}}\right)^{2}-q}}}$

då gäller

${\displaystyle x_{1}+x_{2}=-p\,\!}$
${\displaystyle x_{1}\cdot x_{2}=q\,\!}$

För ${\displaystyle a\geq 0,\ b\geq 0,\ c>0}$:

 ${\displaystyle {\sqrt {a}}\cdot {\sqrt {a}}=a\,\!}$ ${\displaystyle {\sqrt {a}}\cdot {\sqrt {b}}={\sqrt {ab}}\,\!}$ ${\displaystyle b{\sqrt {a}}={\sqrt {b^{2}a}}\,\!}$ ${\displaystyle {\frac {\sqrt {a}}{\sqrt {c}}}={\sqrt {\frac {a}{c}}}\,\!}$ ${\displaystyle {\frac {a}{\sqrt {c}}}={\frac {a{\sqrt {c}}}{c}}\,\!}$ ${\displaystyle {\sqrt[{n}]{ab}}={\sqrt[{n}]{a}}{\sqrt[{n}]{b}}\,\!}$ ${\displaystyle {\sqrt[{n}]{\frac {a}{c}}}={\frac {\sqrt[{n}]{a}}{\sqrt[{n}]{c}}}\,\!}$ ${\displaystyle {\sqrt[{m}]{\sqrt[{n}]{a}}}={\sqrt[{mn}]{a}}\,\!}$ ${\displaystyle a{\sqrt[{n}]{b}}={\sqrt[{n}]{a^{n}b}}\,\!}$ ${\displaystyle {\sqrt[{m}]{\sqrt[{n}]{a}}}={\sqrt[{n}]{\sqrt[{m}]{a}}}\,\!}$ ${\displaystyle {\sqrt[{nq}]{a^{mq}}}={\sqrt[{n}]{a^{m}}}\,\!}$

## Potensregler

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

## Logaritmer

För ${\displaystyle y>0,\ a>0,\ a\neq 1}$:

 ${\displaystyle y=10^{x}\Leftrightarrow x=\log _{10}\ y=\lg \ y\,\!}$ ${\displaystyle y=a^{x}\Leftrightarrow x=\log _{a}\ y\,\!}$ ${\displaystyle y=e^{x}\Leftrightarrow x=\ln \ y\,\!}$ ${\displaystyle \ln \ y=\ln \ 10\cdot \lg \ y\approx 2,3026\ \lg \ y\,\!}$ ${\displaystyle \lg \ y=\lg \ e\cdot \ln \ y\approx 0,4343\ \ln \ y\,\!}$

### Logaritmlagar

 ${\displaystyle a^{\log _{a}x}=x\,\!}$ ${\displaystyle \log(ab)=\log a+\log b\,\!}$ ${\displaystyle a>0\ och\ b>0\,\!}$ ${\displaystyle \log {\frac {a}{b}}=\log a-\log b\,\!}$ ${\displaystyle \log _{a}a^{n}=n\log a\,\!}$ ${\displaystyle \log _{a}{\sqrt[{n}]{a}}={\frac {1}{n}}\log a\,\!}$ ${\displaystyle a^{\frac {\log b}{\log a}}=b\,\!}$