Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\),=> a=bk:c=dk
Ta có : \(\dfrac{ab}{cd}=\dfrac{bkb}{dkd}=\dfrac{kb^2}{kd^2}=\dfrac{b^2}{d^2}\) (1)
\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\) (2)
Từ (1) và (2) => \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\) (đpcm)
Từ \(\dfrac{a}{b}=\dfrac{c}{d}\)
=> \(\dfrac{a}{c}=\dfrac{b}{d}\)
=> Ta sẽ có : \(\dfrac{a}{c}\). \(\dfrac{b}{d}\) = \(\dfrac{ab}{cd}\) = \(\dfrac{a^2}{c^2}\) = \(\dfrac{b^2}{d^2}\) (*1)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\) (*2)
Từ (1);(2) => \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\) (ĐPCM)
