Question

Chứng minh rằng nếu (a² + b² + c²)(x² + y² + z²) = (ax + by + cz)² với x, y, z khác 0 thì \(\dfrac{a}{x}\)=\(\dfrac{b}{y}\)=\(\dfrac{c}{z}\).

Answer 1

Ta có:(a²+b²+c²)(x²+y²+z²)=(ax+by+cz)²

=>a²x²+a²y²+a²z²+b²x²+b²y²+b²z²+c²x²+

c²y²+c²z²=a²x²+b²y²+c²z²+2axby+2axcz+

2bycz

=>a²y²+a²z²+b²x²+b²z²+c²x²+c²y²-2axby-2axcz-2bycz=0

=>(a²y²-2axby+b²x²)+(a²z²-2axcz+c²x²)+

(b²z²-2bycz+c²y²)=0

=>(ay-bx)²+(az-cx)²+(bz-cy)²=0

Vì (ay-bx)²\(\ge0\);(az-cx)²\(\ge0\);(bz-cy)²\(\ge0\)

nên =>(ay-bx)²+(az-cx)²+(bz-cy)²\(\ge0\)

Dấu "=" xảy ra khi:\(\left\{{}\begin{matrix}ay-bx=0\\az-cx=0\\bz-cy=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}ay=bx\\az=cx\\bz=cy\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}\dfrac{a}{x}=\dfrac{b}{y}\\\dfrac{a}{x}=\dfrac{c}{z}\\\dfrac{b}{y}=\dfrac{c}{z}\end{matrix}\right.\)=>\(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)(x;y;z\(\ne0\))