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