1) Giải

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk,c=dk\)

Ta có:

\(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\left(k-1\right)}=\dfrac{k+1}{k-1}\left(1\right)\)

\(\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\left(2\right)\)Từ 1 và 2 suy ra \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\) đpcm

a)\(2^{300}\) và\(3^{200}\)

ta có :

\(2^{300}=2^{3\cdot100}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=3^{2\cdot100}=\left(3^2\right)^{100}=9^{100}\)

Vì 8<9\(\Rightarrow8^{100}< 9^{100}\Rightarrow2^{300}< 3^{200}\)(đpcm)