Câu hỏi của Minh Tú

Question

Cho $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=k$ và a+b+c=abc

Tìm k để $\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=k$

Phương Trâm · Accepted Answer

Ta có:

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=k$

$\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=k^2$

$\Rightarrow\dfrac{1}{a^2}+\dfrac{2}{ab}+\dfrac{1}{b^2}+\dfrac{2}{bc}+\dfrac{1}{c^2}+\dfrac{2}{ac}=k^2$

$\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{1\left(a+b+c\right)}{abc}=k^2$

$\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=k^2-k$Câu trả lời: Ta có:

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=k$

$\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=k^2$

$\Rightarrow\dfrac{1}{a^2}+\dfrac{2}{ab}+\dfrac{1}{b^2}+\dfrac{2}{bc}+\dfrac{1}{c^2}+\dfrac{2}{ac}=k^2$

$\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{1\left(a+b+c\right)}{abc}=k^2$

$\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=k^2-k$

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