P= abc(\(\frac{1}{^{a^3}}\)+\(\frac{1}{b^3}\)+\(\frac{1}{c^3}\)) = abc[(\(\frac{1}{a}\)+\(\frac{1}{b}\))3+\(\frac{1}{c^3}\)-\(\frac{3}{a^2b}\)-\(\frac{3}{ab^2}\)]=abc[(\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\))(....)- \(\frac{3}{a^2b}\)-\(\frac{3}{ab^2}\)]
=abc.(- \(\frac{3}{a^2b}\)-\(\frac{3}{ab^2}\)) =-3(\(\frac{c}{a}\)+\(\frac{c}{b}\)) = -3c(\(\frac{1}{a}\)+\(\frac{1}{b}\)) = -3c.\(\frac{-1}{c}\)=3
P = 3
Đầu tiên,bạn cần chứng minh x + y + z = 0 thì x3 + y3 + z3 = 3xyz ( Bạn ko biết c/m thì hỏi nhé)
Thay\(x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3.\frac{1}{a}.\frac{1}{b}.\frac{1}{c}=\frac{3}{abc}\)
\(\Rightarrow M=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}=\frac{abc}{c^3}+\frac{abc}{a^3}+\frac{abc}{b^3}=abc\left(\frac{1}{c^3}+\frac{1}{a^3}+\frac{1}{b^3}\right)=abc.\frac{3}{abc}=3\)
