Question

puteti sa-mi explicati rezolvarea asta ?
f (x)=^3-2x^2-2x+1
determinati nr a astfel in cat
 1/x1x2+1/x2x3+1/x3x1=a(x1x2+x2x3+x1x3)
unde x1 x2 x3 sunt radacinile ecuatiei f
x1+x2+x3/x1x2x3=a(x1x2+x2x3+x1x3)=2/-1

Answer 1

$\displaystyle Din~relatii~lui~Viete~avem:~ \\ \\ x_1+x_2+x_3=2 \\ \\ x_1x_2+x_2x_3+x_1x_3=-2 \\ \\ x_1x_2x_3=-1. \\ \\ Relatia~\frac{1}{x_1x_2}+ \frac{1}{x_2x_3}+ \frac{1}{x_1x_3}=a(x_1x_2+x_2x_3+x_1x_3)~devine: \\ \\ \frac{x_1+x_2+x_3}{x_1x_2x_3}=a(x_1x_2+x_2x_3+x_1x_3) \Leftrightarrow \frac{2}{-1}=a \cdot (-2) \Leftrightarrow -2=-2a. \\ \\ Deci~a=1.$