Question

cho a,b,c đôi một khác nhau tt/m:

\(a^3+b^3+c^3=3abc\left(abc\ne0\right)\)

chứng minh p=\(\frac{ab^2}{a^2+b^2-c^2}+\frac{bc^2}{b^2+c^2-a^2}+\frac{ca^2}{c^2+a^2-b^2}=0\)

Answer 1

\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+2ab-bc-ca+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Rightarrow a^3+b^3+c^3=3abc\Leftrightarrow a+b+c=0\) vì \(a\ne b\ne c\)

\(\Rightarrow P=\frac{ab^2}{\left(a+b\right)^2-c^2-2ab}+...\)

\(=\frac{ab^2}{\left(a+b+c\right)\left(a+b-c\right)-2ab}+..\)

\(=\frac{ab^2}{-2ab}+\frac{bc^2}{-2bc}+\frac{ca^2}{-2ca}=-\left(\frac{a+b+c}{2}\right)=0\)