a) Ta co: a/b = c/d= k
=> a=bk
c=dk
Ta co: a-b/a+b = bk-b/bk+b = b(k-1)/b(k+1) = k-1/k+1 (1)
Ta co: c-d/c+d = dk-d/dk+d = d(k-1)/d(k+1) = k-1/k+1 (2)
Tu (1) va (2)
=> a-b/a+b=c-d/c+d
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (*)
a) Từ (*) ta có:
\(\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\) (1)
\(\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\) (2)
Từ (1) và (2) suy ra \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
b) Từ (*) ta có:
\(\dfrac{7a-4b}{3a+5b}=\dfrac{7bk-4b}{3bk+5b}=\dfrac{b\left(7k-4\right)}{b\left(3k+5\right)}=\dfrac{7k-4}{3k+5}\) (3)
\(\dfrac{7c-4d}{3c+5d}=\dfrac{7dk-4d}{3dk+5d}=\dfrac{d\left(7k-4\right)}{d\left(3k+5\right)}=\dfrac{7k-4}{3k+5}\) (4)
Từ (3) và (4) suy ra \(\dfrac{7a-4b}{3a+5b}=\dfrac{7c-4d}{3c+5d}\)
c) Từ (*) ta có:
\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\) (5)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) (6)
\(\dfrac{\left(c-a\right)^2}{\left(d-b\right)^2}=\dfrac{\left[\left(dk\right)-\left(bk\right)\right]^2}{\left(d-b\right)^2}=\dfrac{\left[k\left(d-b\right)\right]^2}{\left(d-b\right)^2}=k^2\) (7)
Từ (5), (6) và (7) suy ra \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(c-a\right)^2}{\left(d-b\right)^2}\)
