Selective Question

If a, b, c are rational numbers but not zero, a+b+c=0 , what is the value of

\dfrac{1}{b^2+c^2-a^2} +\dfrac{1}{c^2+a^2-b^2} +\dfrac{1}{a^2+b^2-c^2}

  1. ×

    Positive numbers

  2. ×

    Negative numbers

  3. Zero

  4. ×

    Uncertain

Steven Zheng Steven Zheng posted 3 weeks ago


Answer

  1. \because a+b+c=0

    \therefore (a+b)^2=c^2

    \therefore a^2+b^2-c^2=2ab

    Similarly,

    b^2+c^2-a^2=2bc

    a^2+c^2-a^2=2ac

    \dfrac{1}{b^2+c^2-a^2} +\dfrac{1}{c^2+a^2-b^2} +\dfrac{1}{a^2+b^2-c^2}

    =\dfrac{1}{2ab} +\dfrac{1}{2bc} +\dfrac{1}{2ab}

    =\dfrac{a+b+c}{2abc }

    =0

Steven Zheng Steven Zheng posted 3 weeks ago

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