Question

Cho a , b , c \(\ne\) 0  thõa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) và abc =1

Tình GTBT F \((a^3b^3+b^3c^3+c^3a^3)(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3})\)

Answer 1

Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}=3\) (abc=1) (tự c/m)

Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\)

=>ab+bc=-ca => (ab+bc)³=-c³a³

=>a³b³+b³c³+3a²b².bc+3ab.b²c²=-c³a³

=>a³b³+b³c³+3ab²c(ab+bc)=-c³a³

=>a³b³+b³c³+3ab²c.(-ca)=-c³a³

=>a³b³+b³c³-3a²b²c²=-c³a³

=>a³b³+b³c³+c³a³=3a²b²c² = 3 (do abc=1)

Vậy F=3.3=9