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»NUMERE REALE

Data publicarii: 21 Iulie, 2010

TEORIE

Partea intreaga a unui numar real:

Oricare ar fi numarul real a, exista numarul intreg k, astfel incat a € [k, k+1);

acest numar k intreg, notat [a], se numeste partea intreaga a numarului real a.

Rezulta de aici:

${[a]}\leq{a}<{[a]+1}$ {[a]}\leq{a}<{[a]+1}

${a-1}<{[a]}\leq{a},\;\forall{a}\in{\mathbb{R}}.$ {a-1}<{[a]}\leq{a},\;\forall{a}\in{\mathbb{R}}.

Partea fractionara a unui numar real:

$\{a\}={a-[a]}\Rightarrow{{0}\leq\{a\}<{1}},\;\forall{a}\in{\mathbb{R}}.$ \{a\}={a-[a]}\Rightarrow{{0}\leq\{a\}<{1}},\;\forall{a}\in{\mathbb{R}}.

Puteri:

1) Cu exponent real:

Fie a,b numere pozitive si x,y numere reale oarecare; atunci:

${a^x}\cdot{a^y}=a^{x+y};$ {a^x}\cdot{a^y}=a^{x+y};
$\frac{a^x}{a^y}=a^{x-y};$ \frac{a^x}{a^y}=a^{x-y};
${(\frac{a}{b})}^{x}=\frac{a^x}{b^x};$ {(\frac{a}{b})}^{x}=\frac{a^x}{b^x};
${({a}\cdot{b})}^x={a^x}\cdot{b^x};$ {({a}\cdot{b})}^x={a^x}\cdot{b^x};
${(a^x)}^{y} = {a}^{xy}.$ {(a^x)}^{y} = {a}^{xy}.

2) Cu exponent intreg negativ:

Fie a numar real nenul si n natural nenul.

Se numeste "puterea a (- n) - a a lui a", notata $a^{-n},$ a^{-n},

numarul real definit prin:

$a^{-n}=\frac{1}{a^n}.$ a^{-n}=\frac{1}{a^n}.

Observaţie:

Prin definitie:

$a^{0}=1,\forall{a\neq{0}}.$ a^{0}=1,\forall{a\neq{0}}.

3) Cu exponent rational:

Fie a > 0 si

$r=\frac{m}{n},{m}\in{\mathbb{Z}},\;{n}\in{{\mathbb{N}}^{*}},\;{n}\geq{2}.$ r=\frac{m}{n},{m}\in{\mathbb{Z}},\;{n}\in{{\mathbb{N}}^{*}},\;{n}\geq{2}.

Se numeste "a la puterea r" numarul real notat:

${a^r}={a}^{\frac{m}{n}}=\sqrt[n]{a^m}.$ {a^r}={a}^{\frac{m}{n}}=\sqrt[n]{a^m}.

Radicali de ordinul n (n natural mai mare sau egal cu 2):

1) Daca n este un numar par, mai mare sau egal cu 2 si a nenegativ (mai mare sau

egal cu zero), se numeste radacina de ordinul n din a (sau radacina aritmetica de

ordinul n din a), numarul nenegativ, notat prin

$\sqrt[n]{a},$ \sqrt[n]{a},

astfel incat:

${(\sqrt[n]{a})}^{n} = a.$ {(\sqrt[n]{a})}^{n} = a.

2) Daca n este impar, mai mare sau egal cu 3 si a un numar real oarecare, se numeste

radacina de ordinul n a lui a (sau radacina aritmetica de ordinul n a lui a) numarul

real notat prin

$\sqrt[n]{a},$ \sqrt[n]{a},

astfel incat:

$(\sqrt[n]{a})^{n}=a.$ (\sqrt[n]{a})^{n}=a.

Proprietati ale radicalilor:

Radacina unui produs este egala cu produsul radacinilor:

$\sqrt[n]{{a}\cdot{b}}=\begin{cases}\sqrt[n]{a}\cdot\sqrt[n]{b},\forall{a,b}\geq{0}\\\sqrt[n]{|a|}\cdot\sqrt[n]{|b|},\forall{{a}\cdot{b}}\geq{0}\end{cases};$ \sqrt[n]{{a}\cdot{b}}=\begin{cases}\sqrt[n]{a}\cdot\sqrt[n]{b},\forall{a,b}\geq{0}\\\sqrt[n]{|a|}\cdot\sqrt[n]{|b|},\forall{{a}\cdot{b}}\geq{0}\end{cases};

Radacina unui cat este egala cu catul radacinilor:

$\sqrt[n]{\frac{a}{b}}=\begin{cases}\frac{\sqrt[n]{a}}{\sqrt[n]{b}},\forall{a,b}\geq{0},{b}\neq{0}\\\frac{\sqrt[n]{|a|}}{\sqrt[n]{|b|}},\forall{{a}\cdot{b}}\geq{0},{b}\neq{0}\end{cases};$ \sqrt[n]{\frac{a}{b}}=\begin{cases}\frac{\sqrt[n]{a}}{\sqrt[n]{b}},\forall{a,b}\geq{0},{b}\neq{0}\\\frac{\sqrt[n]{|a|}}{\sqrt[n]{|b|}},\forall{{a}\cdot{b}}\geq{0},{b}\neq{0}\end{cases};

Puterea unei radacini:

${(\sqrt[n]{a})}^{m}=\sqrt[n]{a^m},\forall{a}\ge{0},m\in{{\mathbb{N}}^*};$ {(\sqrt[n]{a})}^{m}=\sqrt[n]{a^m},\forall{a}\ge{0},m\in{{\mathbb{N}}^*};

Simplificarea unei radacini:

$\sqrt[mn]{a^m}=\begin{cases}\sqrt[n]{a}, {a}\geq{0}, {m}\in{{\mathbb{N}}^{*}}\\\sqrt[n]{|a|}, {a}<{0}, {m}\; {pair}\end{cases};$ \sqrt[mn]{a^m}=\begin{cases}\sqrt[n]{a}, {a}\geq{0}, {m}\in{{\mathbb{N}}^{*}}\\\sqrt[n]{|a|}, {a}<{0}, {m}\; {pair}\end{cases};

Radacina unei radacini:

$\sqrt[n]{\sqrt[m]{a}}=\begin{cases}\sqrt[mn]{a},\forall{a}\geq{0}, {m}\in{{\mathbb{N}}^{*}},{m}\geq{2},{n}\in{{\mathbb{N}}^{*}},{n}={2k},{k}\in{{\mathbb{N}}^{*}}\\\sqrt[mn]{a},\forall{a}\in{\mathbb{R}},{m}\in{{\mathbb{N}}^{*}},{m}={2k+1},{n}={2p+1},{k,p}\in{{\mathbb{N}}^{*}}\end{cases};$ \sqrt[n]{\sqrt[m]{a}}=\begin{cases}\sqrt[mn]{a},\forall{a}\geq{0}, {m}\in{{\mathbb{N}}^{*}},{m}\geq{2},{n}\in{{\mathbb{N}}^{*}},{n}={2k},{k}\in{{\mathbb{N}}^{*}}\\\sqrt[mn]{a},\forall{a}\in{\mathbb{R}},{m}\in{{\mathbb{N}}^{*}},{m}={2k+1},{n}={2p+1},{k,p}\in{{\mathbb{N}}^{*}}\end{cases};

Introducerea unui factor sub un radical:

${a}\sqrt[n]{b}=\begin{cases}\sqrt[n]{{a^n}\cdot{b}},\;{a,b}\geq{0},\;{n}={2k},\;{k}\in{{\mathbb{N}}^{*}}\\-\sqrt[n]{{a^n}\cdot{b}},\;{a<0},{b\geq{0}},\;{n}={2k},\;{k}\in{{\mathbb{N}}^{*}}\\\sqrt[n]{{a^n}\cdot{b}},\;\forall{a,b}\in{\mathbb{R}},\;{n}={2p+1},\;{p}\in{{\mathbb{N}}^{*}}\end{cases};$ {a}\sqrt[n]{b}=\begin{cases}\sqrt[n]{{a^n}\cdot{b}},\;{a,b}\geq{0},\;{n}={2k},\;{k}\in{{\mathbb{N}}^{*}}\\-\sqrt[n]{{a^n}\cdot{b}},\;{a<0},{b\geq{0}},\;{n}={2k},\;{k}\in{{\mathbb{N}}^{*}}\\\sqrt[n]{{a^n}\cdot{b}},\;\forall{a,b}\in{\mathbb{R}},\;{n}={2p+1},\;{p}\in{{\mathbb{N}}^{*}}\end{cases};

Scoaterea unui factor de sub un radical:

$\sqrt[n]{{a^n}\cdot{b}}=\begin{cases}{a}\sqrt[n]{b},{a,b}\geq{0},{n}={2k},{k}\in{{\mathbb{N}}^{*}}\\-{a}\sqrt[n]{b},{a}<{0},{b}\geq{0},{n}={2k},{k}\in{{\mathbb{N}}^{*}}\\{a}\cdot\sqrt[n]{b},\forall{a,b}\in{\mathbb{R}},{n}={2p+1},{p}\in{{\mathbb{N}}^{*}}\end{cases}.$ \sqrt[n]{{a^n}\cdot{b}}=\begin{cases}{a}\sqrt[n]{b},{a,b}\geq{0},{n}={2k},{k}\in{{\mathbb{N}}^{*}}\\-{a}\sqrt[n]{b},{a}<{0},{b}\geq{0},{n}={2k},{k}\in{{\mathbb{N}}^{*}}\\{a}\cdot\sqrt[n]{b},\forall{a,b}\in{\mathbb{R}},{n}={2p+1},{p}\in{{\mathbb{N}}^{*}}\end{cases}.

Formula radicalilor compusi:

$\sqrt{a\pm\sqrt{b}}=\sqrt{\frac{a+\sqrt{{a^2}-b}}{2}}\pm\sqrt{\frac{a-\sqrt{{a^2}-b}}{2}},\;{a,b}\geq{0},\;{{a}^{2}-{b}}\geq{0}.$ \sqrt{a\pm\sqrt{b}}=\sqrt{\frac{a+\sqrt{{a^2}-b}}{2}}\pm\sqrt{\frac{a-\sqrt{{a^2}-b}}{2}},\;{a,b}\geq{0},\;{{a}^{2}-{b}}\geq{0}.