Câu hỏi của Nguyễn Kiều Anh

Question

Cho $\left\{{}\begin{matrix}a+b+c=abc\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\a,b,c\ne0\end{matrix}\right.$

Tính $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$

Trần Thanh Phương · Accepted Answer

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2$

$\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4$

$\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4$

$\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\cdot\frac{a+b+c}{abc}=4$

$\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\cdot1=4$

$\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2$

Vậy....Câu trả lời: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2$

$\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4$

$\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4$

$\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\cdot\frac{a+b+c}{abc}=4$

$\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\cdot1=4$

$\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2$

Vậy....

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