Question

[Giỏi Toán 8]

Cho \(\dfrac{a}{b}=\dfrac{c}{d}\)và b, d khác 0. CMR \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

Answer 1

[ILoveMath]

Tham khảo:Chứng minh a/b=c/d hoặc a/b=d/c biết (a^2+b^2)/(c^2+d^2)=ab/cd - An Nhiên

Answer 2

[Nguyễn Tân Vương]

\(\text{Cho }\dfrac{a}{b}=\dfrac{d}{c}\text{ và }b,d\notin0\text{.CMR:}\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

\(\text{Ta có:}\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\text{Lại có:}\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=\dfrac{\left(bd\right).k^2}{bd}=k^2\)

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

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