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 \(\dfrac{m}{n}=\dfrac{p}{q}=k\) ⇒ m=nk ; p=qk

Khi đó,

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

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

Từ (1), (2) ⇒ \(\dfrac{mp}{nq}=\dfrac{m^2+p^2}{n^2+q^2}\)(đpcm)

CHÚC BẠN HỌC TỐT!!!!!!!!!

Answer 2

Đặt \(\dfrac{m}{n}=\dfrac{p}{q}=k=>m=kn,p=qk\)

Ta có \(\dfrac{mp}{nq}=\dfrac{kn.qk}{nq}=\dfrac{k^{2^{ }}\left(nq\right)}{nq}=k^2\left(1\right)\)

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

Từ (1), (2) => ..............

Answer 3

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