Question

[Vương Nguyễn Thanh Triều]

Mọi nguời giúp mình bài này với ạ.

Cho (a²+b²+c²) . (x²+y²+z²) = (ax+by+cz)² và x,y,z khác 0. C/m a/x=b/y=c/z

Answer 1

[Le Thi Khanh Huyen]

\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)

\(\Leftrightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)

\(=a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz\)

Trừ cả hai vế cho \(a^2x^2+b^2y^2+c^2z^2\), có :

\(a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2axby+2bycz+2axcz\)

\(\Rightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2-2axby-2bycz-2axcz=0\)

\(\left(a^2y^2+b^2x^2-2axby\right)+\left(a^2z^2+c^2x^2-2axcz\right)+\left(b^2z^2+c^2y^2-2bycz\right)=0\)

\(\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)

Mà \(\hept{\begin{cases}\left(ay-bx\right)^2\ge0\\\left(az-cx\right)^2\ge0\\\left(bz-cy\right)^2\ge0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}ay-bx=0\\az-cx=0\\bz-cy=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}ay=bx\\az=cx\\bz-cy\end{cases}}\)

\(\Leftrightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)

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