Question

[Nguyễn Ngọc Gia Hân]

Cho a,b,c là các số thực thỏa mãn (a+b+c).(ab+bc+ca)=abc. Tính

P=\(\frac{\left(a+b+c\right)^{2019}}{a^{2019}+b^{2019}+c^{2019}}\)

Answer 1

[Y]

\(\left(a+b+c\right)\left(ab+bc+ca\right)=abc\)

\(\Rightarrow\left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+3abc\right)-abc=0\)

\(\Rightarrow a^2b+bc^2+2abc+a^2c+ac^2+b^2c+ab^2=0\)

\(\Rightarrow b\left(a+c\right)^2+ac\left(a+c\right)+b^2\left(a+c\right)=0\)

\(\Rightarrow\left(a+c\right)\left[b\left(a+c\right)+ac+b^2\right]=0\)

\(\Rightarrow\left(a+c\right)\left(a+b\right)\left(b+c\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a+c=0\Rightarrow a^{2019}+c^{2019}=0\\b+c=0\Rightarrow b^{2019}+c^{2019}=0\\a+b=0\Rightarrow a^{2019}+b^{2019}=0\end{matrix}\right.\)

\(\Rightarrow P=1\)

*Hằng đẳng thức cần áp dụng:

\(x^n+y^n=\left(x+y\right)\left(x^{n-1}-x^{n-2}y+...-xy^{n-2}+y^{n-1}\right)\)

nên \(x+y=0\Rightarrow x^n+y^n=0\)