\(\text { Với } \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3 \text { và } a+b+c=a b c \text { . Tính } \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \text { . }\)

* Hướng dẫn giải

 \(\begin{aligned} &\text { Ta có: } \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3 \Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2}=9 \Leftrightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+2\left(\frac{1}{a b}+\frac{1}{a c}+\frac{1}{b c}\right)=9 \\ &\text { Mà } a+b+c=a b c \Rightarrow \frac{a+b+c}{a b c}=1 \Leftrightarrow \frac{1}{b c}+\frac{1}{a c}+\frac{1}{a b}=1 \\ &\text { Nên } \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+2\left(\frac{1}{a b}+\frac{1}{a c}+\frac{1}{b c}\right)=9 \Leftrightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=9-2=7 \end{aligned}\)