Question

cho \(xy\ne0\)

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

C/m \(\dfrac{a}{x}=\dfrac{b}{y}\)

Answer 1

Do \(xy\ne0\Rightarrow x;y\ne0\)

Ta có : \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

\(\Leftrightarrow a^2x^2+b^2x^2+a^2y^2+b^2y^2=a^2x^2+2axby+b^2y^2\)

\(\Leftrightarrow b^2x^2+a^2y^2=2axby\)

\(\Leftrightarrow b^2x^2+a^2y^2-2axby=0\)

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

Do \(\left(bx-ay\right)^2\ge0\Rightarrow bx-ay=0\)

\(\Rightarrow bx=ay\)

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