Sửa đề: Cho \(\dfrac{a}{b}=\dfrac{c}{d}.CMR:\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}.\)
Giải:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk.\)
Ta có:
\(\dfrac{ab}{cd}=\dfrac{bkb}{dkd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}_{\left(1\right)}.\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}_{\left(2\right)}.\)
Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\Rightarrow\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\left(đpcm\right).\)
Ta có \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=> \(a=bk\\ c=dk\)
Xét \(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^{2\left(1\right)}\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2.d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^{2\left(2\right)}\)
Từ (1) và (2) => \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
