Bài 2:
a: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=b\cdot k;c=d\cdot k\)
\(\dfrac{4a-3b}{a}=\dfrac{4\cdot bk-3b}{bk}=\dfrac{b\left(4k-3\right)}{bk}=\dfrac{4k-3}{k}\)
\(\dfrac{4c-3d}{c}=\dfrac{4\cdot dk-3d}{dk}=\dfrac{d\left(4k-3\right)}{dk}=\dfrac{4k-3}{k}\)
Do đó: \(\dfrac{4a-3b}{a}=\dfrac{4c-3d}{c}\)
b: \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\)
\(\dfrac{3a^2+2b^2}{3c^2+2d^2}=\dfrac{3\cdot\left(bk\right)^2+2b^2}{3\cdot\left(dk\right)^2+2d^2}\)
\(=\dfrac{b^2\left(3k^2+2\right)}{d^2\left(3k^2+2\right)}=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{3a^2+2b^2}{3c^2+2d^2}\)