a) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Khi đó (2a + 3c)(2b - 3d) 

= (2bk + 3dk)(2b - 3d)

= k(2b + 3d)(2b - 3d) (1)

(2a - 3c)(2b + 3d)

= (2bk - 2dk)(2b + 3d)

= k(2b - 3d)(2b + 3d) (2)

Từ (1)(2) => (2a + 3c)(2b - 3d) = (2a - 3c)(2b + 3d)

b) Sửa đề (4a + 3b)(4c - 3d) = (4a - 3b)(4c + 3d) 

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Ta có (4a + 3b)(4c - 3d) = (4bk + 3b)(4dk - 3d) = bd(4k + 3)(4k - 3) (1)

Lại có (4a - 3b)(4c + 3d) = (4bk - 3b)(3dk + 3d) = bd(4k- 3)(4k + 3) (2)

Từ (1)(2) => (4a + 3b)(4c - 3d) = (4a - 3b)(4c + 3d) 

1, Ta có: \(\frac{a}{b}=\frac{c}{d}\)

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

\(\Rightarrow\left(2a+3c\right).\left(2b-3d\right)=\left(2a-3c\right).\left(2b+3d\right)\)

        Vậy (2a + 3c).(2b - 3d) = (2a - 3c).(2b + 3d)

Câu 2 cũng tương tự nên tự làm đi

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=bk;c=dk\)

1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)

\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)

Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)

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

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

Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)

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

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

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

4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)

\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)

Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)

Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\left(a+c\right)\cdot\left(b-d\right)=\left(bk+dk\right)\left(b-d\right)=k\left(b^2-d^2\right)\)

\(\left(a-c\right)\left(b+d\right)=\left(bk-dk\right)\left(b+d\right)=k\left(b^2-d^2\right)\)

Do đó: \(\left(a+c\right)\left(b-d\right)=\left(a-c\right)\left(b+d\right)\)

b: \(\left(2a+3c\right)\left(2b-3d\right)=\left(2bk+3dk\right)\left(2b-3d\right)=k\left(4b^2-9d^2\right)\)

\(\left(2a-3c\right)\left(2b+3d\right)=\left(2bk-3dk\right)\left(2b+3d\right)=k\left(4b^2-9d^2\right)\)

Do đó: \(\left(2a+3c\right)\left(2b-3d\right)=\left(2a-3c\right)\left(2b+3d\right)\)

đặt a/b =c/d =k 

=> a=bm , c=dm 

=> 2a+3c/2b+3d =2bm+3bm/ 2b +3d = m.(2d+3d)/2d+3d =m (1)

=> 2a-3c/2d-3d=2bm-3dm /2b -3d =m.(2b-3d)/2b-3d= m (2)

Từ (1) và (2) => 2a+3c/2b+3d =2a-3c/2b-3d 

câu 2 tương tự nha

bạn khôi đặt là k mà lại khi m

sai đề r, a/3 là s, phải a/b chứ, nếu là a/b thì lm ntnày:

Lấy a/b=c/d=k(k thuộc N*) 
=>a=bk ; c=dk 
Xét : + 2a-3c/2b-3d=2bk-3dk/2b-3d= k^2.(2b-3d)/2b-3d=k^2 (1) 
       + 2a+3c/2b+3d=2bk+3dk/2b+3d= k^2.(2b+3d)/2b+3d=k^2 (2) 
(1);(2)=> 2a-3c/2b-3d=2a+3c/2b+3d(đpcm)

Vậy 2a-3c/2b-3d=2a+3c/2b+3d

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

\(\Rightarrow a=bk\); \(c=dk\)

Ta có: \(\frac{2a+c}{2b+d}=\frac{2bk+dk}{2b+d}=\frac{k\left(2b+d\right)}{2b+d}=k\)(1)

\(\frac{2a-3c}{2b-3d}=\frac{2bk-3dk}{2b-3d}=\frac{k\left(2b-3d\right)}{2b-3d}=k\)(2)

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