Question Number 161408 by mathlove last updated on 17/Dec/21
Commented by som(math1967) last updated on 17/Dec/21
$$\mathrm{9} \\ $$
Answered by behi834171 last updated on 17/Dec/21
![12=[Σa][Σ(1/a)]=3+(a/b)+(a/c)+(b/a)+(b/c)+(c/a)+(c/b)= =3+Σ((a^2 +b^2 )/(ab))⇒Σ ((a^2 +b^2 )/(ab))=12−3=9 .■](https://www.tinkutara.com/question/Q161438.png)
$$\mathrm{12}=\left[\Sigma{a}\right]\left[\Sigma\frac{\mathrm{1}}{{a}}\right]=\mathrm{3}+\frac{{a}}{{b}}+\frac{{a}}{{c}}+\frac{{b}}{{a}}+\frac{{b}}{{c}}+\frac{{c}}{{a}}+\frac{{c}}{{b}}= \\ $$$$=\mathrm{3}+\Sigma\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{ab}}\Rightarrow\Sigma\:\frac{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} }{\boldsymbol{{ab}}}=\mathrm{12}−\mathrm{3}=\mathrm{9}\:.\blacksquare \\ $$