đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
a)
\(\frac{5a+2c}{5b+2d}=\frac{5bk+2dk}{5b+2d}=\frac{k\left(5b+2d\right)}{5b+2d}=k\)
\(\frac{a-4c}{b-4d}=\frac{bk-4dk}{b-4d}=\frac{k\left(b-4d\right)}{b-4d}=k\)
=>\(\frac{5a+2c}{5b+2d}=\frac{a-4c}{b-4d}=k\)(đpcm)
b)
\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}=\frac{b}{d}\)
\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a+b}{c+d}=\frac{bk+b}{dk+d}=\frac{b\left(k+1\right)}{d\left(k+1\right)}=\frac{b}{d}\)
=>\(\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)