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}
Positive numbers
Negative numbers
Zero
Uncertain
Steven Zheng
posted 4 months ago
\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
posted 4 months ago