Question

Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\)

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

Answer 1

vì \(\frac{a}{b}=\frac{c}{d}\)=>\(\frac{ac}{bd}=\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)

=>đpcm

Answer 2

Cách khác:

Dat  \(\frac{a}{b}=\frac{c}{d}=k\) thì \(a=bk;c=dk\)\(\Rightarrow\frac{ac}{bd}=\frac{bk.dk}{bd}=k^2\left(1\right)\)

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

Từ \(\left(1\right),\left(2\right)\Rightarrow\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)

Answer 3

Vì \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{ac}{bd}=\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)

\(\Rightarrowđpcm\)

Answer 4

\(\text{Đặt}\)\(\frac{a}{b}=\frac{c}{d}=k\)\(\text{thì}\)\(a=bk;c=dk\)

\(\text{Ta có:}\)\(\frac{ac}{bd}=\frac{bk.dk}{bd}=\frac{bd.k^2}{bd}=k^2\)\(\text{(1)}\)

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

\(\text{Từ (1) và (2) suy ra}\)\(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)