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

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

\(\Rightarrow dpcm\)

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

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

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

Suy ra \(\dfrac{a}{b}=\dfrac{c}{d}\)

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

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

Áp dụng dãy tỉ số bằng nhau ta có:

\(\frac{2A+3C}{2B+3D}=\frac{2A-3C}{2B-3D}=\frac{2A+3C+2A-3C}{2B+3D+2B-3D}=\frac{4A}{4B}=\frac{A}{B}\left(1\right)\)\(\frac{2A+3C}{2B+3D}=\frac{2A-3C}{2B-3D}=\frac{2A+3C-2A+3C}{2B+3D-2B+3D}=\frac{6C}{6D}=\frac{C}{D}\left(2\right)\)

Từ (1) và (2) suy ra : \(\frac{A}{B}=\frac{C}{D}\)

Giải :

Từ đảng thức : \(\frac{2a+3c}{2b+3d}=\frac{2a-3c}{2b-3d}\)

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

\(\Rightarrow4ab-6ad+6bc-9cd=4ab-6bc+6ad-9cd\)

\(\Rightarrow\left(4ab-6ad+6bc-9cd\right)-\left(4ab-6bc+6ad-9cd\right)=0\)

\(\Rightarrow4ab-6ad+6bc-9cd-4ab+6bc-6ad+9cd=0\)

\(\Rightarrow\left(4ab-4ab\right)-\left(6ad+6ad\right)+\left(6bc+6bc\right)-\left(9cd-9cd\right)=0\)

\(\Rightarrow-12ad+12bc=0\)

\(\Rightarrow12bc=12ad\)

\(\Rightarrow bc=ad\)

\(\Rightarrow\frac{a}{b}=\frac{c}{d}\left(\text{đpcm}\right)\)

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

Vì \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=kd\)

\(\Rightarrow\frac{2a-3c}{2b-3d}=\frac{2bk-3dk}{2b-3d}=\frac{k\left(2b-3d\right)}{2b-3d}=k\)(1)

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

\(\RightarrowĐPCM\)