Question

Chứng minh rằng nếu \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\) với x,y khác 0 thì \(\dfrac{a}{x}=\dfrac{b}{y}\)

Answer 1

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

\(\Rightarrow\left(ax\right)^2+\left(ay\right)^2+\left(bx\right)^2+\left(by\right)^2=\left(ax\right)^2+2.ax.by+\left(by\right)^2\)

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

\(\Rightarrow\left(ay-bx\right)^2=0\Rightarrow ay-bx=0\Rightarrow ay=bx\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}\)

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