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

a) => \(\frac{2a+c}{2b+d}=\frac{2kb+kd}{2b+d}=\frac{k\left(2b+d\right)}{2b+d}=k\) (1)

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

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

b) => \(\frac{ab}{cd}=\frac{kbb}{kdd}=\frac{b^2}{d^2}\) (1)

\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(kb\right)^2+b^2}{\left(kd\right)^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\) (2)

Từ (1) và (2) => \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)

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

Áp dụng .... ta có:

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

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

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

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

Vậy \(\frac{a}{b}=\frac{c}{d}\)

a)

i) Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{b}{a}=\frac{d}{c}.\)

\(\Rightarrow\frac{b}{a}+1=\frac{d}{c}+1\)

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

\(\Rightarrow\frac{b+a}{a}=\frac{d+c}{c}.\)

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

Chúc bạn học tốt!

còn ii và phần b nữa

Lời giải:

a)

Đặt $\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt, c=dt$

i. Khi đó:

$\frac{a}{a+b}=\frac{bt}{bt+b}=\frac{bt}{b(t+1)}=\frac{t}{t+1}(1)$

$\frac{c}{c+d}=\frac{dt}{dt+d}=\frac{dt}{d(t+1)}=\frac{t}{t+1}(2)$

Từ $(1);(2)\Rightarrow \frac{a}{a+b}=\frac{c}{c+d}$ (đpcm)

ii.

$\frac{a-b}{c-d}=\frac{bt-b}{dt-d}=\frac{b(t-1)}{d(t-1)}=\frac{b}{d}(3)$

$\frac{a+b}{c+d}=\frac{bt+b}{dt+d}=\frac{b(t+1)}{d(t+1)}=\frac{b}{d}(4)$

Từ $(3);(4)\Rightarrow \frac{a-b}{c-d}=\frac{a+b}{c+d}$ (đpcm)

b)

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

$\Leftrightarrow 2ac-4ad+bc-2bd=2ac+ad-4bc-2bd$

$\Leftrightarrow 5bc=5ad\Leftrightarrow bc=ad\Leftrightarrow \frac{a}{b}=\frac{c}{d}$

Ta có đpcm.

Lời giải:

a)

Đặt $\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt, c=dt$

i. Khi đó:

$\frac{a}{a+b}=\frac{bt}{bt+b}=\frac{bt}{b(t+1)}=\frac{t}{t+1}(1)$

$\frac{c}{c+d}=\frac{dt}{dt+d}=\frac{dt}{d(t+1)}=\frac{t}{t+1}(2)$

Từ $(1);(2)\Rightarrow \frac{a}{a+b}=\frac{c}{c+d}$ (đpcm)

ii.

$\frac{a-b}{c-d}=\frac{bt-b}{dt-d}=\frac{b(t-1)}{d(t-1)}=\frac{b}{d}(3)$

$\frac{a+b}{c+d}=\frac{bt+b}{dt+d}=\frac{b(t+1)}{d(t+1)}=\frac{b}{d}(4)$

Từ $(3);(4)\Rightarrow \frac{a-b}{c-d}=\frac{a+b}{c+d}$ (đpcm)

b)

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

$\Leftrightarrow 2ac-4ad+bc-2bd=2ac+ad-4bc-2bd$

$\Leftrightarrow 5bc=5ad\Leftrightarrow bc=ad\Leftrightarrow \frac{a}{b}=\frac{c}{d}$

Ta có đpcm.

ta có : ab=cd⇔ad=bc⇔4ad=4bc⇔2ad+2ad=2bc+2bcab=cd⇔ad=bc⇔4ad=4bc⇔2ad+2ad=2bc+2bc

⇔2ad−2bc=2bc−2ad⇔ac+2ad−2bc−4bd=ac+2bc−2ad−4bd⇔2ad−2bc=2bc−2ad⇔ac+2ad−2bc−4bd=ac+2bc−2ad−4bd

⇔(c+2d)(a−2b)=(a+2b)(c−2d)⇔a+2bc+2d=a−2bc−2d(đpcm) 

Bạn ơi! Phải chứng minh \(\frac{a}{b}=\frac{c}{d}\) chứ!

a) 

i) theo đề ta có ad=bc

ta có a(c+d) = ac+ad

ta có (a+b)c = ac+bc

mà ad = bc

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

các bạn ơi mình không hiểu sao câu ii mình ra thế này

 ii) đặt \(\frac{a}{b}=\frac{c}{d}=m\)\(\Rightarrow\)a=mb ; c=dm

Ta có \(\frac{a-b}{c-d}\)= \(\frac{mb-b}{md-d}\)=\(\frac{b\left(m-1\right)}{d\left(m-1\right)}\)=\(\frac{b}{d}\)

Ta có \(\frac{a+c}{b+d}\)=\(\frac{mb+md}{b+d}\)=m

\(\frac{a}{a+2b}=\frac{c}{c+2d}\Rightarrow ac+2ad=ac+2bc\Rightarrow2ad=2bc\Rightarrow bc=ad\Rightarrow\frac{a}{b}=\frac{c}{d}\)

\(\frac{b}{2a-b}=\frac{d}{2c-d}\Rightarrow2cb-bd=2ad-bd\Rightarrow2ad=2cb\Rightarrow ad=cd\Rightarrow\frac{a}{b}=\frac{c}{d}\)