Question

Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) Chứng minh rằng \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

Answer 1

Đặt :

\(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow a=bk;c=dk\)

\(VT=\dfrac{ac}{bd}=\dfrac{bkdk}{bd}=\dfrac{bdk^2}{bd}=k^2\left(1\right)\)

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

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

Answer 2

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) =>\(a=bk,c=dk\)

=> \(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k.k=k^2\left(1\right)\)

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

=\(\dfrac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)

Từ (1)và(2)=>\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

Chúc Bạn Học Tốt

Answer 3

Đặt \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) = k \(\Rightarrow\) a = bk; c = dk

\(\Rightarrow\) + \(\dfrac{ac}{bd}\) = \(\dfrac{bk.dk}{bd}\) = k . k = k² (1)

+ \(\dfrac{a^2+c^2}{b^2+d^2}\) = \(\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}\)= \(\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}\) = \(\dfrac{k^2.\left(b^2+d^2\right)}{b^2+d^2}\) = k² (2)

Từ (1) và (2) => \(\dfrac{ac}{bd}\) = \(\dfrac{a^2+c^2}{b^2+d^2}\)

Answer 4

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{ac}{bd}\left(1\right)\)

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

Từ \(\left(1\right)\) và \(\left(2\right)\) ta có:

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

\(\Rightarrowđpcm\)