Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk,c=dk\)
a) Ta có:
\(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2}{d^2}\) (1)
\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\frac{b^2}{d^2}\) (2)
Từ (1) và (2) suy ra \(\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Vậy \(\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b) Ta có:
\(\frac{a^4+b^4}{c^4+d^4}=\frac{\left(bk\right)^4+b^4}{\left(dk\right)^4+d^4}=\frac{b^4.k^4+b^4}{d^4.k^4+d^4}=\frac{b^4.\left(k^4+1\right)}{d^4.\left(k^4+1\right)}=\frac{b^4}{d^4}\) (1)
\(\frac{\left(a+b\right)^4}{\left(c+d\right)^4}=\frac{\left(bk+b\right)^4}{\left(dk+d\right)^4}=\frac{\left[b\left(k+1\right)\right]^4}{\left[d\left(k+1\right)\right]^4}=\frac{b^4}{d^4}\) (2)
Từ (1) và (2) suy ra \(\frac{a^4+b^4}{c^4+d^4}=\frac{\left(a+b\right)^4}{\left(c+d\right)^4}\)
Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)
\(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)(đpcm)
