Question

cho \(\dfrac{a}{b}=\dfrac{c}{d}\) , cmr: \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

Answer 1

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\). \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có:

\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=\dfrac{bdk^2}{bd}=k^2\)

\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)

\(\Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\left(=k^2\right)\)

\(\Rightarrowđpcm\)

Answer 2

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow a=n.c,b=n.d\)

Thế vô rồi làm nha!

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