Question

Chứng minh rằng nếu:

\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2;x,y\ne0\) thì \(\dfrac{a}{x}=\dfrac{b}{y}\)

Answer 1

Ta có: (a² + b²)(x² + y²)

= (ax)² + a²y² + b²x² + (by)²

= (ax + by)² - 2abxy + a²y² + b²x²

= (ax + by)² + (a²y² + b²x² - 2abxy)

Mà (a² + b²)(x² + y²) = (ax + by)²

\(\Rightarrow\) a²y² + b²x² - 2abxy = 0

\(\Rightarrow\) \(\left(ay\right)^2-2.ay.bx+\left(bx\right)^2=0\)

\(\Rightarrow\) \(\left(ay-bx\right)^2=0\)

\(\Rightarrow\) \(ay=bx\)

\(\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}\) (đpcm)