Daniela Andreea  in data de 10.02 a trimis :

0,(16)+2x-0,(3)=4-2(x+1)+3,(6)

Rezolvare pentru ecuatia :

0,\left(16\right)+2x-0,\left(3\right)=4-2\left(x+1\right)+3,\left(6\right)

Ca sa rezolvam ecuatia transformam mai intai fractiile periodice mixte in fractii ordinare, astfel obtinem

\frac{16}{99}+2x-\frac{3}{9}=4-2x-2+\frac{36-3}{9}\Rightarrow    \frac{1\cdot 16+99\cdot 2x-11\cdot 3}{99}=2-2x+\frac{33}{9}

aducem acum la acelasi numitor si in partea dreapta

\frac{1\cdot 16+99\cdot 2x-11\cdot 3}{99}=\frac{9\cdot 2-9\cdot 2x+1\cdot 33}{9}\Rightarrow    \frac{16+198x-33}{99}=\frac{18-18x+33}{9}

Acum folosim teorema fundamantala a proportiilor, adica produsul mezilor este egal cu produsul extremilor

\frac{198x-17}{99}=\frac{21-18x}{9}\Rightarrow 9\left(198x-17\right)=99\left(51-18x\right)|:9\Rightarrow    198x-17=11\left(51-18x\right)\Rightarrow 198x-17=561-198x\Rightarrow 198x+198x=561+17\Rightarrow 396x=578\Rightarrow x=\frac{578}{396}^{(2}=\frac{289}{198}

Acum efectuam proba:

\frac{16}{99}+2x-\frac{3}{9}=4-2x-2+\frac{33}{9}\Rightarrow    \frac{16}{99}+2\cdot\frac{289}{198}-\frac{3}{9}=2-2\cdot \frac{289}{198}+\frac{33}{9}\Rightarrow    \frac{16}{99}+\frac{289}{99}-\frac{3}{9}=2-\frac{289}{99}+\frac{33}{9}\Rightarrow    \frac{16+289-11\cdot 3}{99}=\frac{99\cdot 2-289+11\cdot 33}{99}\Rightarrow    \frac{305-33}{99}=\frac{198-289+363}{99}\Rightarrow    \frac{272}{99}=\frac{272}{99}

 

Maria in data de 10.02 a trimis :

Calculati :

2\sqrt{3}\left(\sqrt{3} + \sqrt{2}\right) + 3\sqrt{2}\left(\sqrt{2} - \sqrt{3}\right)

Ca sa rezolvam exercitiul de mai sus efectuam produsul numarului din fata parantezei cu fiecare termen, acelasi lucru il efectuam si la cea de-a doua paranteza, astfel obtinem:

2\sqrt{3}\left(\sqrt{3} + \sqrt{2}\right) + 3\sqrt{2}\left(\sqrt{2} - \sqrt{3}\right)=

2\sqrt{3}\cdot \sqrt{3}+2\sqrt{3}\cdot \sqrt{2}+3\sqrt{2}\cdot\sqrt{2}-3\sqrt{2}\cdot\sqrt{3}=

2\sqrt{3\cdot 3}+2\sqrt{3\cdot 2}+3\sqrt{2\cdot 2}-3\sqrt{2\cdot 3}=

2\sqrt{3^{2}}+2\sqrt{6}+3\sqrt{2^{2}}-3\sqrt{6}=2\cdot 3+2\sqrt{6}+3\cdot 2-3\sqrt{6}=

6+2\sqrt{6}+6-3\sqrt{6}=12-\sqrt{6}

Observati ca dupa ce am efectuat produsul radicalilor, am folosit regulile de calcul cu radicali, iar apoi am efectuat calculele, adica am adunat, respectiv am scazut teremenii asemenea, am adunat numerele naturale 6+6=12, dar am si scazut termenii care care au avut radicali astfel am obtinut :

2\sqrt{6}-3\sqrt{6}=\sqrt{6}\left(2-3\right)=\sqrt{6}\left(-1\right)=-\sqrt{6}.

Karina 11.02

2) Scrie numerele din 20 in 20 de la 10 la 100.

10; 30; 50; 70; 90

Mamuleanu in data de 11.02

Aflati un numar daca jumatatea acestuia marita cu \frac{11}{6} este egala cu \frac{47}{15}

Solutie :

Notam cu x numarul, acum formam ecuatia

\frac{x}{2}+\frac{11}{6}=\frac{47}{15}

Rezolvam ecuatia

\frac{x}{2}=\frac{47}{15}-\frac{11}{6}\Rightarrow \frac{x}{2}=\frac{2\cdot 47-5\cdot 11}{30}\Rightarrow \frac{x}{2}=\frac{94-55}{30}\Rightarrow \frac{x}{2}=\frac{39}{30}^{(3}\Rightarrow \frac{x}{2}=\frac{13}{10}\Rightarrow x=\frac{2\cdot 13}{10}\Rightarrow x=\frac{26}{10}

Deci numarul este \frac{26}{10}

Sau daca simplificam acum prin 2, obtinem \frac{26}{10}^{(2}=\frac{13}{5}

Efectuam proba:

\frac{x}{2}+\frac{11}{6}=\frac{47}{15}    \\ \frac{\frac{26}{10}}{2}+\frac{11}{6}=\frac{26}{10}\cdot\frac{1}{2}+\frac{11}{6}=\frac{26}{20}+\frac{11}{6}=\frac{3\cdot 26+10\cdot 11}{60}=\frac{78+110}{60}=\frac{188}{60}^{(4}=\frac{47}{15}

 

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