a )\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=\frac{2a}{2c}\)

\(\frac{a-b}{c-d}=\frac{2a}{2c}\Rightarrow\frac{a-b}{2a}=\frac{c-d}{2c}\) ( đpcm)

b ) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{3a}{3c}=\frac{2b}{2d}=\frac{5a-3b}{5c-3d}=\frac{3a+2b}{3c+2d}\)

\(\Rightarrow\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}\) ( đpcm )

a) \(\hept{\begin{cases}\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{5a-3b}{5c-3d}\\\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{3a}{3c}=\frac{2b}{2d}=\frac{3a+2b}{3c+2d}\end{cases}}\)

\(\Rightarrow\frac{5a-3b}{5c-3d}=\frac{3a+2b}{3c+2d}\)

\(\Rightarrow\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}\)

b) Chứng minh tương tự 

ko biet nghen

\(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}\)

Áp dụng dãy tỉ số bằng nhau ta có:

\(\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{5a-3b}{5c-3d}\)

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

=>\(\frac{5a-3b}{5c-3d}=\frac{a}{c}=\frac{3a+2b}{3c+3d}\)

=>\(\frac{5a-3b}{5c-3d}=\frac{3a+2b}{3c+3d}\)

=>\(\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+3d}\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{5a-3b}{5c-3d}\)

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

=> \(\frac{5a-3b}{5c-3d}=\frac{3a+2b}{3c+2d}\) ( Vì cùng bằng \(\frac{a}{c}\))

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\)\(\frac{a}{c}=\frac{b}{d}\)\(\Rightarrow\)\(\frac{5a}{5c}=\frac{3b}{3d}=\frac{3a}{3c}=\frac{2b}{2d}\)

=> \(\frac{5a-3b}{5c-3d}=\frac{3a+2b}{3c+2d}\Rightarrow\)\(\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}\) (đpcm)

          \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\)

          \(\dfrac{a}{c}\) = \(\dfrac{b}{d}\)

   \(\dfrac{a}{c}\)  =  \(\dfrac{5a}{5c}\) = \(\dfrac{3b}{3d}\) Áp dụng tính chất dãy tỉ số bằng nhau ta có:

      \(\dfrac{a}{c}\) =   \(\dfrac{5a+3b}{5c+3d}\) (1) 

       \(\dfrac{a}{c}\) = \(\dfrac{5a-3b}{5c-3d}\)  (2)

Kết hợp (1) và (2) ta có:

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

⇒   \(\dfrac{5a+3b}{5a-3b}\) =  \(\dfrac{5c+3d}{5c-3d}\) (đpcm)

b;   \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) 

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

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

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

a.a/b=c/d=>.a/c=b/d=>2a/2c=b/d

ap dung tính chất dãy tỉ sồ bàng nhau ya có

2a/2c=b/d=2a+b/2c+d=2a-b/2c-d

=>2a+b/2a-b=2c+d/2c-d

b.a/b=c/d=>a/c=b/d=>5a/5c=3b/3d=3a/3c=2b/2d

áp dụng  tính chat dãy ti số bang nhau ta co

5a/5c=3b/3d=3a/3c=2b/2d=5a-3b/5c-3d=3a+2b/3c+2d

5a-3b/3a+2b=5c-3d/3c+2d

bạn bấm vào đây cho mình nhé !CMR:từ tỉ lệ thức $\frac{a}{b}=\frac{c}{d}$ab =cd  ta suy ra được $\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}$5a−3b3a+2b =5c−3d3c+2d 

dat a/b=c/d=k                                           a=kb          ;c=kd                                               Xet 5a+3b/5a-3b=5kb+3b/5kb-3b= b(5k+3)/b(5k-3)=5k+3/5k-3    (1)                    Xet 5c+3d/5c-3d=5kd+3d/5kd-3d=                d(5k+3)/d(5k-3)= 5k+3/5k-3   (2)                   Tu (1) va (2) Suy ra 5a+3b/5a-3b =5c+3d/5c-3d