Cho a;b;c;x;y;z thỏa mãn
\(\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\)
Tính gía trị \(\dfrac{x^{2017}+y^{2017}+z^{2017}}{2016}\)Hung nguyen
\(\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\)
\(\Leftrightarrow x^2\left(\dfrac{1}{a^2}-\dfrac{1}{a^2+b^2+c^2}\right)+y^2\left(\dfrac{1}{b^2}-\dfrac{1}{a^2+b^2+c^2}\right)+z^2\left(\dfrac{1}{c^2}-\dfrac{1}{a^2+b^2+c^2}\right)=0\)
\(\Leftrightarrow\dfrac{x^2\left(b^2+c^2\right)}{a^2}+\dfrac{y^2\left(a^2+c^2\right)}{b^2}+\dfrac{z^2\left(a^2+b^2\right)}{c^2}=0\)
Vì \(a,b,c\ne0\)nên dấu = xảy ra khi \(x=y=z=0\)
Vậy \(\dfrac{x^{2017}+y^{2017}+z^{2017}}{2016}=0\)
