Question

Cho \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)Tính giá trị của biểu thức N= \(\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}\)

Answer 1

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Answer 2

Ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^3=0\)

\(\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)

Ta lại có:

\(N=\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}\)

\(\Leftrightarrow N=abc.\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=abc.\dfrac{3}{abc}=3\)