Question

cho \(\frac{a}{b}=\frac{c}{d}\)cm \(\left(\frac{a-b}{c-d}\right)^2=\frac{a.b}{c.d}\)

Answer 1

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

Ta có: \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\frac{\left[b.\left(k-1\right)\right]^2}{\left[d.\left(k-1\right)\right]^2}=\frac{b^2.\left(k-1\right)^2}{d^2.\left(k-1\right)^2}=\frac{b^2}{d^2}\)

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

=>\(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{b^2}{d^2}=\frac{a.b}{c.d}\)

=>\(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{a.b}{c.d}\)