Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=>a=bk; c=dk
a: \(\frac{a-2b}{b}=\frac{bk-2b}{b}=\frac{b\left(k-2\right)}{b}=k-2\)
\(\frac{c-2d}{d}=\frac{dk-2d}{d}=\frac{d\left(k-2\right)}{d}=k-2\)
Do đó: \(\frac{a-2b}{b}=\frac{c-2d}{d}\)
b: \(\frac{a-2c}{3a+c}=\frac{bk-2\cdot dk}{3\cdot bk+dk}=\frac{k\left(b-2d\right)}{b\left(3b+d\right)}=\frac{b-2d}{3b+d}\)
c: \(\frac{a^2-2b^2}{\left(a+4b\right)^2}=\frac{\left(bk\right)^2-2b^2}{\left(bk+4b\right)^2}=\frac{b^2\left(k^2-2\right)}{b^2\left(k+4\right)^2}=\frac{k^2-2}{\left(k+4\right)^2}\)
\(\frac{c^2-2d^2}{\left(c+4d\right)^2}=\frac{\left(dk\right)^2-2\cdot d^2}{\left(dk+4d\right)^2}=\frac{d^2\left(k^2-2\right)}{d^2\left(k+4\right)^2}=\frac{k^2-2}{\left(k+4\right)^2}\)
Do đó: \(\frac{a^2-2b^2}{\left(a+4b\right)^2}=\frac{c^2-2d^2}{\left(c+4d\right)^2}\)
d: \(\left(a+4c\right)\left(2b-3d\right)\)
\(=\left(bk+4\cdot dk\right)\left(2b-3d\right)=k\left(b+4d\right)\left(2b-3d\right)\) (2)
\(\left(b+4d\right)\left(2a-3c\right)\)
\(=\left(b+4d\right)\left(2\cdot bk-3\cdot dk\right)=k\left(2b-3d\right)\left(b+4d\right)\) (1)
Từ (1),(2) suy ra (a+4c)(2b-3d)=(b+4d)(2a-3c)
e: \(\frac{ac}{bd}=\frac{bk\cdot dk}{bd}=k^2\)
\(\frac{a^2-c^2}{b^2-d^2}=\frac{\left(bk\right)^2-\left(dk\right)^2}{b^2-d^2}=\frac{k^2\left(b^2-d^2\right)}{b^2-d^2}=k^2\)
Do đó: \(\frac{ac}{bd}=\frac{a^2-c^2}{b^2-d^2}\)
