một trong 3 câu đầy đủ nha
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{k-1}{k+1}\)
\(\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{k-1}{k+1}\)
Do đó: \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
b: \(\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{2k+5}{3k-4}\)
\(\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{2k+5}{3k-4}\)
Do đó: \(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
c: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
