Lời giải:

Đặt $\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk$.

Khi đó:

$\frac{5a+3b}{5a-3b}=\frac{5bk+3bk}{5bk-3bk}=\frac{8bk}{2bk}=4(1)$

$\frac{5c+3d}{5c-3d}=\frac{5dk+3dk}{5dk-3dk}=\frac{8dk}{2dk}=4(2)$

Từ $(1); (2)$ suy ra điều phải chứng minh.

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

Suy ra \(\begin{cases}a=bk\\c=dk\end{cases}\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\Leftrightarrow\frac{bk+b}{b}=\frac{dk+d}{d}\)

Xét VT \(\frac{bk+b}{b}=\frac{b\left(k+1\right)}{b}=k+1\left(1\right)\)

Xét VP \(\frac{dk+d}{d}=\frac{d\left(k+1\right)}{d}=k+1\left(2\right)\)

Từ (1) và (2) -->Đpcm

b)Đặt tương tự ta có:

\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\Leftrightarrow\frac{5bk+3b}{5bk-3b}=\frac{5dk+3d}{5dk-3d}\)

Xét VT \(\frac{5bk+3b}{5bk-3b}=\frac{b\left(5k+3\right)}{b\left(5k-3\right)}=\frac{5k+3}{5k-2}\left(1\right)\)

Xét VP \(\frac{5dk+3d}{5dk-3d}=\frac{d\left(5k+3\right)}{d\left(5k-3\right)}=\frac{5k+3}{5k-3}\left(2\right)\)

Từ (1) và (2) -->Đpcm

Bạn xem lại đề nhé :)

1) Ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\)

2) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{5}{3}.\frac{a}{b}=\frac{5}{3}.\frac{c}{d}\Rightarrow\frac{5a}{3b}-1=\frac{5c}{3d}-1\Rightarrow\frac{5a-3b}{3b}=\frac{5c-3d}{3d}\)

\(\Rightarrow\frac{3b}{5a-3b}=\frac{3d}{5c-3d}\Rightarrow\frac{6b}{5a-3b}=\frac{6d}{5c-3d}\Rightarrow\frac{6b}{5a-3b}+1=\frac{6d}{5c-3d}+1\)

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

1) Vì a/b = c/d

=> a/b + 1 = c/d + 1

=> a + b/b = c + d/d (đpcm)

2) Vì a/b = c/d

=> a/c = b/d

=> 5a/5c = 3b/3d = 5a + 3b/5c + 3d = 5a - 3b/5c - 3d ( theo tc DTSBN )

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

1,a/b=c/d

=>\(\frac{a}{b}+1=\frac{c}{d}+1\)

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

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

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

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

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\left(k\ne0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5.bk+3b}{5.bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)(1)

\(\dfrac{5c+3d}{5c-3d}=\dfrac{5.dk+3d}{5.dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)(2)

Từ (1) và (2) \(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)