Question

Cho biết: \(\dfrac{m}{n}=\dfrac{p}{q}\) . Chứng minh rằng\(\dfrac{mp}{nq}=\dfrac{m^2+p^2}{n^2+q^2}\)

Answer 1

Đặt \(\dfrac{m}{n}=\dfrac{p}{q}=k\Rightarrow\left\{{}\begin{matrix}m=nk\\p=qk\end{matrix}\right.\)

\(\Rightarrow\dfrac{mp}{nq}=\dfrac{nk.qk}{nq}=k^2\left(1\right)\)

\(\Rightarrow\dfrac{m^2+p^2}{n^2+q^2}=\dfrac{\left(nk\right)^2+\left(qk\right)^2}{n^2+q^2}=\dfrac{n^2k^2+q^2k^2}{n^2+q^2}=\dfrac{k^2\left(n^2+q^2\right)}{n^2+q^2}=k^2\left(2\right)\)

Từ (1) và (2) suy ra: \(\dfrac{mp}{nq}=\dfrac{m^2+p^2}{n^2+q^2}\left(đpcm\right)\)