Giải chi tiết

Áp dụng định lí Vi-ét ta có: x1+x2=10x1x2=−8 

Ta có x12+x22=x1+x22−2x1x2=102−2−8=116

Khi đó

A=1x12+1x22=x12+x22x1x22=116−82=2916

 C=x1−x2x12−x22=x1−x2x1−x2x1+x2=x1−x22x1+x2

     =x12+x22−2x1x2x1+x2=116−2.−8.10=1320

D=x1−x2⇒D2=x1−x22=x12+x22−2x1x2⇒D2=132⇒D=233(Vì D>0)

E=x14+x24

x14+x24+2x12x22−2x12x22=x12+x222−2x1x22=1162−2.−82=13328         

F=x15+x25=x13+x23=x1+x23−3x1x2x1+x2=103−3.10(−8)=1240

F=x15+x25=x12+x22x13+x23−x12x23−x13x22

=x12+x22x13+x23−x12x22x1+x2=116.1240−−82.10=143200