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

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

Như vậy, \(\frac{{a - c}}{{b - d}} = \frac{{a + 2c}}{{b + 2d}}\) (đpcm)

Lời giải:

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

Ta có:

$(a+2c)(b+d)=(bk+2dk)(b+d)=k(b+2d)(b+d)(1)$

$(a+c)(b+2d)=(bk+dk)(b+2d)=k(b+d)(b+2d)(2)$

Từ $(1); (2)\Rightarrow (a+2c)(b+d)=(a+c)(b+2d)$

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

\(\Leftrightarrow ab+2ad-2bc-4cd=ab+2bc-2ad-4cd\)

\(\Leftrightarrow2ad+2ad=2bc+2bc\Leftrightarrow4ab=4bc\)

\(\Leftrightarrow ad=bc\Rightarrow\dfrac{a}{b}=\dfrac{c}{d},\left(a,b,c,d\ne0\right)\)

1) Ta có:
\(\dfrac{a}{a+b}\)=\(\dfrac{c}{c+d}\)
=>a.(c+d) = c.(a+b)
a.c+a.d = a.c+b.d
Do đó a.d=b.d
=>\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)( đpcm)

Câu 2: 

Đặt a/b=c/d=k

=>a=bk; c=dk

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

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

=>\(\dfrac{3a+2c}{3b+2d}=\dfrac{-5a+3c}{-5b+3d}\)

b: \(\dfrac{a^2}{b^2}=\dfrac{b^2k^2}{b^2}=k^2\)

\(\dfrac{2c^2-ac}{2d^2-bd}=\dfrac{c\left(2c-a\right)}{d\left(2d-b\right)}=\dfrac{dk}{d}\cdot\dfrac{2dk-bk}{2d-b}=k^2\)

=>\(\dfrac{a^2}{b^2}=\dfrac{2c^2-ac}{2d^2-bd}\)

Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{2a+b}{2a-b}=\dfrac{2bk+b}{2bk-b}=\dfrac{2k+1}{2k-1}\)

\(\dfrac{2c+d}{2c-d}=\dfrac{2dk+d}{2dk-d}=\dfrac{2k+1}{2k-1}\)

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

b: \(\dfrac{2a+b}{a-2b}=\dfrac{2bk+b}{bk-2b}=\dfrac{2k+1}{k-2}\)
\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{2k+1}{k-2}\)

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

\(\frac{a}{b}=\frac{c}{d}=\frac{7a}{7b}=\frac{2c}{2d}=\frac{7a+2c}{7b+2d}\left(1\right)\)  (Tính chất dãy tỷ số bằng nhau)

\(\frac{a}{b}=\frac{c}{d}=\frac{7a}{7b}=\frac{2c}{2d}=\frac{7a-2c}{7b-2d}\left(2\right)\) (lý do như trên)

Từ (1) và (2) \(\Rightarrow\frac{7a+2c}{7b+2d}=\frac{7a-2c}{7b-2d}\)

a/b=c/d=k 

=> a=bk, c=dk

thế vào các biểu thức đó rồi sử dụng phân phối

\(a)\)

Ta có: \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow3a3b=\frac{2c}{2d}\)

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

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

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