Question

[Wendy]

Cho a, b, c \(\ne\)0 thỏa mãn \(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)

Tính P = x²⁰¹⁰ + y²⁰¹¹ + z²⁰¹² + \(\frac{11}{2011}\)

Answer 1

[Nguyệt]

\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\Leftrightarrow\frac{x^2+y^2+z^2}{a^2+b^2+c^2}-\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}=0\)

\(\Leftrightarrow\left(\frac{x^2}{a^2+b^2+c^2}-\frac{x^2}{a^2}\right)+\left(\frac{y^2}{a^2+b^2+c^2}-\frac{y^2}{b^2}\right)+\left(\frac{z^2}{a^2+b^2+c^2}-\frac{z^2}{c^2}\right)=0\)

\(\Leftrightarrow x^2.\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}\right)+y^2.\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}\right)+z^2.\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}\right)=0\)

vì \(a,b,c\ne0\Rightarrow\hept{\begin{cases}\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}\right)\ne0\\\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}\right)\ne0\\\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}\right)\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x^2=0\\y^2=0\\z^2=0\end{cases}}\Rightarrow x=y=z=0\Rightarrow P=0+\frac{11}{2011}=\frac{11}{2011}\)