Operatii cu numere reale

alex52 Dum Oct 28, 2012 6:07 pm

1.Calculati:
radical din [8/7+9/14+10/21+...+250/2401)-(1+1/2+1/3+...+1/343)]
/-supra

2.Calculati:
radical din [(1/1*4)+(1/2*6)+(1/3* Cool

+...+(1/49*48)+(1/49*100)]=
*-inmultit

**ati** Lun Oct 29, 2012 11:32 pm

$\sqrt{(\frac{8}{1*7}+\frac{9}{2*7}+\frac{10}{3*7}+...+\frac{250}{343*7})-(\underbrace{\frac{7}{1*7}+\frac{7}{2*7}+\frac{7}{3*7}+...+\frac{7}{343*7}}_{am\;amplificat\;fiecare\; termen\; cu\; 7})}=\\ =\sqrt{(\frac{8}{1*7}-\frac{7}{1*7})+(\frac{9}{2*7}-\frac{7}{2*7})+(\frac{10}{3*7}-\frac{7}{3*7})+...+(\frac{250}{343*7}-\frac{7}{343*7})}=\\ =\sqrt{\underbrace{1/7+1/7+1/7+...+1/7}_{de\;242\; ori}+243/(343*7)}=\sqrt{\frac{243*343+243}{7}}= ... \;\;\;etc.$

in loc de 250 nu trebuia sa fie 350 Question

**ati** Mar Oct 30, 2012 2:31 am

alex52 a scris:2.Calculati:
radical din [( 1/1*4 )+( 1/2*6 )+( 1/3*8 )+...+( 1/48*98 )+( 1/49*100 )]=
*-inmultit

Aceste tipuri de exercitii de obicei se reduc la utilizarea formulei: 1/[k*(k+1)]=1/k - 1/(k+1)
Sub radical avem ca termen fractiile:
$\frac{1}{1*4}=\frac{1}{2}\cdot\frac{1}{1*2}=1/2\cdot(\frac{1}{1}-\frac{1}{2})\\ \frac{1}{2*6}=1/2\cdot\frac{1}{2*3}=1/2\cdot(\frac{1}{2}-\frac{1}{3})\\ \frac{1}{3*8}=1/2\cdot\frac{1}{3*4}=1/2\cdot(\frac{1}{3}-\frac{1}{4})\\ .....................................................\\ \frac{1}{49*100}=1/2\cdot\frac{1}{49*50}=1/2\cdot(\frac{1}{49}-\frac{1}{50})\\ Se\;face\;suma\;termenilor\;si\;dupa aliminarea \;tremenilor\;asemenea\;opuse\;vom\;avea\\ S=1/2\cdot(\frac{1}{1}-\frac{1}{50})=\frac{1}{2}\cdot\frac{49}{50}=49/25\;\Rightarrow\;\sqrt{S}=7/5\surd\surd$

Continut sponsorizat

Operatii cu numere reale

Operatii cu numere reale

Re: Operatii cu numere reale

Re: Operatii cu numere reale

Re: Operatii cu numere reale