Question

Cho x,y,a,b thỏa mãn 

\(\frac{x^2+y^2}{a^2+b^2}\)= \(\frac{x^2}{a^2}\)+\(\frac{y^2}{b^2}\),a,b\(\ne\)0

Chứng minh x=y=0

Answer 1

\(\dfrac{x^2+y^2}{a^2+b^2}=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\)

\(\Leftrightarrow\dfrac{x^2+y^2}{a^2+b^2}=\dfrac{x^2b^2+a^2y^2}{a^2b^2}\)

\(\Leftrightarrow\left(x^2+y^2\right)a^2b^2=\left(a^2+b^2\right)\left(x^2b^2+a^2y^2\right)\)

\(\Leftrightarrow a^2b^2x^2+a^2b^2y^2=a^2x^2b^2+a^4y^2+b^4x^2+a^2y^2b^2\)

\(\Leftrightarrow0=a^4y^2+b^4x^2\)

Có \(\left\{{}\begin{matrix}a^4y^2\ge0\\b^4x^2\ge0\end{matrix}\right.\) =>\(a^4y^2+b^4x^2\ge0\)

 [=] xảy ra <=> \(\left\{{}\begin{matrix}a^4y^2=0\\b^4x^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\) (vì a;b khác 0)

Vậy y=x=0 (đpcm)