Question

Cho:\(\dfrac{1}{a}\)+\(\dfrac{1}{b}\)+\(\dfrac{1}{c}\)=0.Tính S=\(\dfrac{bc}{a^2}\)+\(\dfrac{ca}{b^2}\)+\(\dfrac{ab}{c^2}\)

Answer 1

a/d vào công thức a^3+b^3+b^3=3abc( khi a+b+c=0)

ta đc 1/a+1/b+1/c=0

=> (1/a)^3+(1/b)^3+(1/c)^3=3. (1/abc)

lại có S=\(\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}\)

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

=3.\(\dfrac{abc}{abc}\)=1

chúc bạn học tốt ^ ^

Answer 2

Dễ CM : nếu x+y+z=0 thì x^3+y^3+z^3=3xyz

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

\(S=\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)\\ =abc.\dfrac{1}{abc}=1\)