Đặt \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\Rightarrow x=ak;y=bk;z=ck\)
Ta có:
\(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left[\left(ak\right)^2+\left(bk\right)^2+\left(ck\right)^2\right]\left(a^2+b^2+c^2\right)=k^2\left(a^2+b^2+c^2\right)^2\)
\(\left(ax+by+cx\right)^2=\left(a^2k+b^2k+c^2k\right)^2=k^2\left(a^2+b^2+c^2\right)^2\)
=> đpcm
\(\left\{{}\begin{matrix}\dfrac{x}{a}=\dfrac{y}{b}\\\dfrac{x}{a}=\dfrac{z}{c}\\\dfrac{y}{b}=\dfrac{z}{c}\end{matrix}\right.\) \(\Rightarrow\) \(\left\{{}\begin{matrix}ay=bx\\az=cx\\bz=cy\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}\left(ay-bx\right)^2=0\\\left(az-cx\right)^2=0\\\left(bz-cy\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\) \(\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)
\(a^2y^2-2axby+b^2x^2+a^2z^2-2axcz+c^2x^2+b^2z^2-2bycz+c^2y^2=0\)
\(a^2y^2+b^2x^2+a^2z^2+c^2x^2+b^2z^2+c^2y^2-\left(2axby+2bycz+2axcz\right)=0\)
\(a^2y^2+b^2x^2+a^2z^2+c^2x^2+b^2z^2+c^2y^2+a^2x^2+b^2y^2+c^2z^2-\left(a^2x^2+b^2y^2+c^2z^22axby+2bycz+2axcz\right)=0\)
\(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)-\left(ax+by+cz\right)^2=0\)
\(\Rightarrow\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\) ( đpcm )
Mệt chết :VV
