Răspuns:
Explicație pas cu pas:
a) x²+mx+2=0, dupa relatiile Viete avem, [tex]\left \{ {{x_{1}+x_{2}=-m} \atop {x_{1}x_{2}=2}} \right.[/tex]
[tex](x_{1}+x_{2})^{2}-2x_{1}x_{2}=5,~deci~(-m)^{2}-2*2=5,~m^{2}=9,~deci~m=-3~sau~m=3.[/tex]
b) x²+mx+m+2=0, dupa relatiile Viete avem, [tex]\left \{ {{x_{1}+x_{2}=-m} \atop {x_{1}x_{2}=m+2}} \right. ~2x_{1}x_{2}=x_{1}+x_{2},~deci~2*(m+2)=-m,~2m+4=-m,~2m+m=-4,~3m=-4,~m=-\frac{4}{3}=-1\frac{1}{3}.[/tex]
c) x²+x+m=0, dupa relatiile Viete avem, [tex]\left \{ {x_{1}+x_{2}=-1} \atop {x_{1}x_{2}=m}} \right. ~\frac{1}{x_{1}+1} +\frac{1}{x_{2}+1}=-\frac{3}{4},~\frac{x_{2}+1+x_{1}+1}{(x_{1}+1)(x_{2}+1)}= -\frac{3}{4},~\frac{x_{1}+x_{2}+2}{x_{1}x_{2}+x_{1}+x_{2}+1} =-\frac{3}{4}\\\frac{-1+2}{m-1+1} =-\frac{3}{4},~\frac{1}{m} =-\frac{3}{4},~m=-\frac{4}{3}=-1\frac{1}{3}.[/tex]
