a/b=c/d=k 

=> a=bk, c=dk

thế vào các biểu thức đó rồi sử dụng phân phối

\(a)\)

Ta có: \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow3a3b=\frac{2c}{2d}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{3a}{3b}=\frac{2c}{2d}=\frac{3a+2c}{3b+2d}\)

\(\Leftrightarrow\frac{3a}{3b}=\frac{3a+2c}{3b+2d}\)hay \(\frac{a}{b}=\frac{3a+2c}{3b+2d}\)

1: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=b\cdot k;c=d\cdot k\)

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

\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\)

Do đó: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)

2: \(\dfrac{2a+b}{a-2b}=\dfrac{2\cdot bk+b}{bk-2b}=\dfrac{b\left(2k+1\right)}{b\left(k-2\right)}=\dfrac{2k+1}{k-2}\)

\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{d\left(2k+1\right)}{d\left(k-2\right)}=\dfrac{2k+1}{k-2}\)

Do đó: \(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)

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

\(\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\)

Do đó: \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)

4: \(\dfrac{5a+3b}{5c+3d}=\dfrac{5\cdot bk+3b}{5dk+3d}=\dfrac{b\left(5k+3\right)}{d\left(5k+3\right)}=\dfrac{b}{d}\)

\(\dfrac{5a-3b}{5c-3d}=\dfrac{5\cdot bk-3b}{5\cdot dk-3d}=\dfrac{b\left(5k-3\right)}{d\left(5k-3\right)}=\dfrac{b}{d}\)

Do đó: \(\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)