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 \(\frac{a}{b}=\frac{c}{d}=k\)

=> a = b.k ; c= d.k

\(\frac{2a+3b}{2a-3b}=\frac{2.\left(b.k\right)+3.b}{2.\left(b.k\right)-3b}=\frac{2b.k+3b}{2b.k-3b}=\frac{2b.\left(k+1,5\right)}{2b.\left(k-1,5\right)}=\frac{k+1,5}{k-1,5}\left(1\right)\)

\(\frac{2c+3d}{2c-3d}=\frac{2.\left(d.k\right)+3d}{2.\left(d.k\right)-3d}=\frac{2d.k+3d}{2d.k-3d}=\frac{2d.\left(k+1,5\right)}{2d.\left(k-1,5\right)}=\frac{k+1,5}{k-1,5}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) => đpcm

bài này đúng không bạn

a/c=c/d suy ra 2a/2c=3b/3d = (2a+3b)/(2c+3d)=(2a-3d)/(2c-3d)

                                       suy ra (2a+3b)/(2a-3b)=(2c+3d)/(2c-3d)

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

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

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

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:

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

d ở đâu ra vậy?