Exercitii rezolvate cu operatii numere reale

Test

Operatii cu numere reale

1. Scrieti sub forma de fractie zecimala numarul $\frac{17}{12}$ .

2. Demonstrati ca numarul $a=\left(\sqrt{8}+5-\frac{4}{\sqrt{2}}\right)\cdot\sqrt{3\frac{6}{25}}$ este patrat perfect.

3. Calculati :

a) $\left(-\frac{5}{2}\right)^{-1}-2^{-1}\cdot\frac{1}{5}+\frac{0,2}{\sqrt{2}}\cdot\sqrt{8}$

b) $\left(\frac{8}{\sqrt{3}}+\frac{7}{\sqrt{2}}\right)\cdot\sqrt{6}-\sqrt{24}\cdot\left(\frac{6}{\sqrt{2}}+\frac{4}{\sqrt{3}}\right)$

4. Fie numerele:

$a=\sqrt{3+2\sqrt{2}}$ si $b=\sqrt{3-2\sqrt{2}}$

a) Aratati ca $a\cdot b=1$

b) Aratati ca $a^{2}-2ab+b^{2}=4$

c) Diferenta $\frac{1}{a}-\frac{1}{b}$ este numar natural.

5) Determinati multimea

$A=\left\{n\in Z|\frac{\sqrt{7+4\sqrt{3}}-\sqrt{5-2\sqrt{6}}+\sqrt{11-6\sqrt{2}}}{3n+1}\in Z\right\}$

6) Daca $E\left(x,y\right)=6x+3y+1$ .

Calculati:

$E\left(a, b\right)$ , unde: $a=\frac{3}{\sqrt{5}+\sqrt{2}}-\left(\sqrt{5}-\sqrt{2}\right)+2^{-1}-3^{-2}\cdot\left(-0,(3)\right)^{-2}$

si $b=\left(\frac{1}{2\sqrt{5}}+\frac{2}{\sqrt{5}}+\frac{3\sqrt{5}}{5}\right)\cdot\left(-\frac{33}{2\sqrt{2}}\right)$

Solutie:

1. Ca sa transformam fractia ordinara in fractie zecimala calculam

$17:12=1,41(6)$

2. Pentru a arata ca numarul a este patrat perfect efectuam calculele in exercitiu, astfel:

$a=\left(\sqrt{8}+5-\frac{4}{\sqrt{2}}\right)\cdot\sqrt{3\frac{6}{25}}$

Mai intai scoatem factorii de sub radicali dar si rationalizam pe unde exista aceasta posibilitate, cat si introducem intregii in fractii, astfel a devine:

$a=\left(\sqrt{2^{2}\cdot 2}+5-\frac{4\cdot\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\right)\cdot\sqrt{\frac{2\cdot 25+6}{25}}$

$a=\left(2\sqrt{2}+5-\frac{4\sqrt{2}}{2}^{(2}\right)\cdot\sqrt{\frac{81}{25}}$

$a=\left(2\sqrt{2}+5-2\sqrt{2}\right)\cdot\frac{\sqrt{81}}{\sqrt{21}}$

Acum efectuam calculele in paranteza rotunda si obtinem:

$a=5\cdot\frac{9}{5}$

Adica $a=\frac{5\cdot 9}{5}=\frac{45}{5}=9$

Deci am obtinut a=9, care este patrat perfect.

3. a) La acest exercitiu trebuie sa efectuam calculele, adica:

$\left(-\frac{5}{2}\right)^{-1}-2^{-1}\cdot\frac{1}{5}+\frac{0,2}{\sqrt{2}}\cdot\sqrt{8}$ dar mai intai ne reamintim ca $a^{-1}=\frac{1}{a}$ , dar si $a^{-n}=\frac{1}{a^{n}}$

Astfel obtinem $-\frac{1}{\frac{5}{2}}-\frac{1}{2}\cdot\frac{1}{5}+\frac{\frac{2}{10}}{\sqrt{2}}\cdot\sqrt{8}$

Acum unde avem factori sub radicali, scoatem factorii de sub radicali si obtinem:

$-\frac{1}{1}:\frac{5}{2}-\frac{1\cdot 1}{2\cdot 5}+\left(\frac{2}{10}:\frac{\sqrt{2}}{1}\right)\cdot\sqrt{2^{2}\cdot 2}= -\frac{1}{1}\cdot\frac{2}{5}-\frac{1}{10}+\frac{2}{10}\cdot\frac{1}{\sqrt{2}}\cdot 2\sqrt{2}=-\frac{2}{5}-\frac{1}{10}+\frac{1}{5}\cdot\frac{1}{\sqrt{2}}\cdot 2\sqrt{2}=-\frac{2}{5}-\frac{1}{10}+\frac{1}{5\sqrt{2}}\cdot 2\sqrt{2}^{(\sqrt{2}}= -\frac{2}{5}-\frac{1}{10}+\frac{1}{5}\cdot 2=-\frac{2}{5}-\frac{1}{10}+\frac{2}{5}=-\frac{1}{10}$

b) $\left(\frac{8}{\sqrt{3}}+\frac{7}{\sqrt{2}}\right)\cdot\sqrt{6}-\sqrt{24}\cdot\left(\frac{6}{\sqrt{2}}+\frac{4}{\sqrt{3}}\right)$

La acest exercitiu mai intai rationalizam si astfel obtinem:

$\left(\frac{8\sqrt{3}}{3}+\frac{7\sqrt{2}}{2}\right)\cdot\sqrt{6}-\sqrt{2^{2}\cdot 6}\cdot\left(\frac{6\sqrt{2}}{2}+\frac{4\sqrt{3}}{3}\right)=$

$\left(\frac{2\cdot 8\sqrt{3}}{6}+\frac{3\cdot 7\sqrt{2}}{6}\right)\cdot\sqrt{6}-2\sqrt{6}\cdot \left(3\sqrt{2}+\frac{4\sqrt{3}}{3}\right)=$

$\left(\frac{16\sqrt{3}}{6}+\frac{21\sqrt{2}}{6}\right)\cdot3\sqrt{2}-2\sqrt{6}\cdot\left(\frac{9\sqrt{2}}{3}+\frac{4\sqrt{3}}{3}\right)=$

$\frac{16\sqrt{3}+21\sqrt{2}}{6}\cdot\sqrt{6}-2\sqrt{6}\cdot\frac{9\sqrt{2}+4\sqrt{3}}{3}$

Acum mai efectuam produsul intre numar si fractie:

$\frac{16\sqrt{3}\cdot\sqrt{6}+21\sqrt{2}\cdot\sqrt{6}}{6}-\frac{2\sqrt{6}\cdot 9\sqrt{2}-2\sqrt{6}\cdot 4\sqrt{3}}{3}=\frac{16\cdot 3\sqrt{2}+21\cdot 2\sqrt{3}}{6}-\frac{18\cdot 2\sqrt{3}-8\cdot 3\sqrt{2}}{3}= \frac{48\sqrt{2}+42\sqrt{3}}{6}-\frac{36\sqrt{3}-24\sqrt{2}}{3}=\frac{6\left(8\sqrt{2}+7\sqrt{3}\right)}{6}-\frac{6\left(6\sqrt{3}+4\sqrt{2}\right)}{3}=8\sqrt{2}+7\sqrt{3}-2\left(6\sqrt{3}+4\sqrt{2}\right)=8\sqrt{2}+7\sqrt{3}-12\sqrt{3}-8\sqrt{2}=-5\sqrt{3}$

4. Pentru a calcula produsul numerelor a si b, mai intai, avem $a=\sqrt{3+2\sqrt{2}}=\sqrt{\left(1+\sqrt{2}\right)^{2}}=|1+\sqrt{2}|=1+\sqrt{2}$

Fie folosim formulele radicalilor complexi, fie formulele de calcul prescurtat.

$b=\sqrt{3-2\sqrt{2}}=\sqrt{\left(1-\sqrt{2}\right)^{2}}=|1-\sqrt{2}|=\sqrt{2}-1$ comutam termenii intre ei, deoarece radicalul obtinut este negativ.

Astfel avem $a\cdot b=1$

b) Stim ca $a^{2}-2ab+b^{2}=\left(a-b\right)^{2}=\left[1+\sqrt{2}-\left(\sqrt{2}-1\right)\right]^{2}=\left(1+sqrt{2}-\sqrt{2}+1\right)^{2}=2^{2}=4$

c) Diferenta $\frac{1}{a}-\frac{1}{b}=\frac{b-a}{a\cdot b}=\frac{\sqrt{2}-1-\left(1+\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\cdot\left(\sqrt{2}-1\right)}=\frac{\sqrt{2}-1-1-\sqrt{2}}{1-2}=\frac{-2}{-1}=2\in N$ .