Question

Ajutor
Din relațiile lui viet

Answer 1

S1=x1+x2=-b/a=-5
S2=x1*x2=c/a=-6

a) [tex] \frac{1}{ x_{1} ^{3}} + \frac{1}{ x_{2} ^{3} }= \frac{ x_{1} ^{3}+ x_{2} ^{3}}{ x_{1} ^{3}* x_{2} ^{3}} [/tex]
Acum va trebui sa aflat cat este numaratorul:
$x_{1} ^{3}+ x_{2} ^{3}=(x_{1}+x_{2}) ^{3} -3x_{1}x_{2}(x_{1}+x_{2}) =S1^{3}-3*S2*S1$
Deci $\frac{ x_{1} ^{3}+ x_{2} ^{3}}{ x_{1} ^{3}* x_{2} ^{3}} =\frac{S1^{3}-3*S2*S1}{S2^{3}}$

Aici inlocuiesti sumele date si obtii raspunsul final

b) 
[tex] \frac{1}{ x_{1} ^{2} -x_{1}+1} + \frac{1}{ x_{2} ^{2} -x_{2}+1}= \frac{ x_{2} ^{2} -x_{2}+1 + x_{1} ^{2} -x_{1}+1}{ (x_{1} ^{2} -x_{1}+1)( x_{2} ^{2} -x_{2}+1)}=[/tex]
$\frac{ (x_{2} +x_{1} )^{2}-2x_{1}x_{2} -(x_{2} +x_{1})+2}{ (x_{1} ^{2} -x_{1}+1)( x_{2} ^{2} -x_{2}+1)}=$
[tex]\frac{ (x_{2} +x_{1} )^{2}-2x_{1}x_{2} -(x_{2} +x_{1})+2} { x_{1} ^{2}x_{2} ^{2} -x_{1}x_{2}(x_{1}+x_{2})+(x_{1}+x_{2})^{2}-x_{1}x_{2}-(x_{1}+x_{2})^{2}+1}=[/tex]
[tex]\frac{ S1^{2}-2S2 -S1)+2} { S2 ^{2} -S2S1+S1^{2}-S2-S1^{2}+1}[/tex]

Aici inlocuiesti sumele date si obtii raspunsul final