a,b,c numere reale pozitive

Stefy Vin Sept 14, 2012 3:26 am

Dacă a, b, c numere reale pozitive, astfel încât $(a+b+c)^{2}=8c(a+b-c)$ , aratati ca ecuatia : 2(ax-c)=c-2bx are solutia 0,5.

**Admin** Vin Sept 14, 2012 3:58 am

Rezolvarea este destul de simpla:
$\begin{array}{l} 2\left( {ax - c} \right) = c - 2bx \Rightarrow x = \frac{{3c}}{{2\left( {a + b} \right)}}\\ {\left( {\underbrace {a + b}_t + c} \right)^2} = 8c\left( {a + b - c} \right) \Leftrightarrow {\left( {t + c} \right)^2} = 8c\left( {t - c} \right)\\ {t^2} + 2tc + {c^2} - 8tc + 8{c^2} = 0 \Leftrightarrow {t^2} - 6tc + 9{c^2} = {\left( {t - 3c} \right)^2} = {\left( {a + b - 3c} \right)^2} = 0 \Rightarrow c = \frac{{a + b}}{3}\\ x = \frac{{3c}}{{2\left( {a + b} \right)}} = \frac{{3 \times \frac{{a + b}}{3}}}{{2\left( {a + b} \right)}} = \frac{1}{2} = 0.5 \end{array}$

a,b,c numere reale pozitive

a,b,c numere reale pozitive

Re: a,b,c numere reale pozitive