Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\left(1\right)\)

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

Từ (1) và (2) suy ra \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3b}\left(=\dfrac{2k+3}{2k-3}\right)\)

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

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

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)\(\Rightarrow a=bk;c=dk\)

Suy ra: \(\frac{2a+3b}{2a-3b}=\frac{2.bk+3b}{2.bk-3b}=\frac{b.\left(2k+3\right)}{b.\left(2k-3\right)}=\)\(\frac{2k+3}{2k-3}\)

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

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

Ta có:\(\frac{a}{b}=\frac{c}{d}\)=>\(\frac{a}{c}=\frac{b}{d}\)=>\(\frac{2a}{2c}=\frac{3b}{3d}\)

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

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

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

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

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

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

Ta có: \(\dfrac{a-b}{c-d}=\dfrac{bk-b}{dk-d}=\dfrac{b}{d}\)

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

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

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

\(\Rightarrow\frac{a}{2a-3b}=\frac{b.k}{2b.k-3b}=\frac{b.k}{\left(2k-3\right)b}=\frac{k}{2k-3}\left(1\right)\)

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

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

Giải:

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

Thay vào vế trái ta có:

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

Thay vào vế phải ta có:

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

\(\Rightarrow VP=VT=\dfrac{2k+3}{2k-3}\Rightarrow\) Đpcm

Ta có :

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

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

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

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

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

(+) \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\)

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

\(\Leftrightarrow4ac+6bc-6ad-9bd=4ac-6bc+6ad-9bd\)

\(\Leftrightarrow12bc=12ad\Leftrightarrow bc=ad\) (đúng)

Vậy (*) đúng (đpcm)

Sửa đề: Chứng minh \(\dfrac{ab}{cd}=\left(\dfrac{2a+3b}{2c+3d}\right)^2\)

Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)

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

Do đó: \(\dfrac{ab}{cd}=\left(\dfrac{2a+3b}{2c+3d}\right)^2\)