Đặt \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\Rightarrow\left\{{}\begin{matrix}x=ak\\y=bk\\z=ck\end{matrix}\right.\)
Ta có:
\(\left\{{}\begin{matrix}\left(a^2k^2+b^2k^2+c^2k^2\right)\left(a^2+b^2+c^2\right)\\\left(a.ak+b.bk+c.ck\right)^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}k^2\left(a^2+b^2+c^2\right)^2\\\left(a^2k+b^2k+c^2k\right)^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}k^2\left(a^2+b^2+c^2\right)^2\\\left[k\left(a^2+b^2+c^2\right)\right]^2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}k^2\left(a^2+b^2+c^2\right)^2\\k^2\left(a^2+b^2+c^2\right)^2\end{matrix}\right.\)
\(\Rightarrow\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)
Vậy......................(đpcm)
Chúc bạn học tốt!!!
