Ta có:

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

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

\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\left(1\right).\)

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

Từ (1) và (2) \(\Rightarrow\frac{a}{b}=\frac{a+c}{b+d}=\frac{a-c}{b-d}\left(đpcm\right).\)
 

Từ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a+c}{b+c}=\frac{a-c}{b-d}\)( tính chất dãy tỉ số bằng nhau )

Theo đề ra, ta có:

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

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

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

a, Áp dụng t/c dtsbn:

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

b, Áp dụng t/c dtsbn:

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{5b}{5d}=\dfrac{3a}{4c}=\dfrac{4b}{4d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)

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

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

\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\)

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

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

Ta có \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)

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

Do đó \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

A) \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt,c=dt\)

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

suy ra đpcm. 

\(\frac{a-b}{c-d}=\frac{bt-b}{dt-d}=\frac{b}{d},\frac{a+b}{c+d}=\frac{bt+b}{dt+d}=\frac{b}{d}\)

suy ra đpcm. 

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

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

suy ra đpcm. 

Lần sau bạn cho thêm cả dấu ngoặc cho dễ hiểu nhé :v

Đặt \(\frac{a}{b}=\frac{c}{d}=k\) => \(\left\{{}\begin{matrix}a=b.k\\c=d.k\end{matrix}\right.\) \(\left(b,d\ne0\right)\)

Thay \(\left\{{}\begin{matrix}a=b.k\\c=d.k\end{matrix}\right.\) vào \(\frac{a^2-b^2}{ab}\) và \(\frac{c^2-d^2}{cd}\) ta có :

\(\left\{{}\begin{matrix}\frac{\left(b.k\right)^2-b^2}{b.k.b}\\\frac{\left(d.k\right)^2-d^2}{d.k.d}\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\frac{b^2.k^2-b^2}{b^2.k}\\\frac{d^2.k^2-d^2}{d^2.k}\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\frac{b^2\left(k^2-1\right)}{b^2.k}\\\frac{d^2\left(k^2-1\right)}{d^2.k}\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}\frac{k^2-1}{k}\\\frac{k^2-1}{k}\end{matrix}\right.\)(vì b,d khác 0 nên \(b^2,d^2\) khác 0)

=> \(\frac{a^2-b^2}{ab}\) = \(\frac{c^2-d^2}{cd}\) (vì cùng bằng \(\frac{k^2-1}{k}\))

vậy \(\frac{a^2-b^2}{ab}\) = \(\frac{c^2-d^2}{cd}\) nếu \(\frac{a}{b}=\frac{c}{d}\)

lâu lắm không làm nên không chắc đâu :v

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>a=bk; c=dk

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

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

Do đó: VT=VP(đpcm)

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

Do đó \(\frac{a+b}{a-b}=\frac{bk+b}{bk-b}=\frac{b\left(k+1\right)}{b\left(k-1\right)}=\frac{k+1}{k-1}\)(1)

\(\frac{c+d}{c-d}=\frac{dk+d}{dk-d}=\frac{d\left(k+1\right)}{d\left(k-1\right)}=\frac{k+1}{k-1}\)(2)

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

Áp dụng tính chất dãy tỉ số bằng nhau

ta có :

a/b=c/d

=>a/c=b/d

áp dụng tính chất dãy tỉ số = nhau ta có:

a/c=b/d=a+b/c+d (1)

a/c=b/d=a-b/c-d (2)

từ 1 và 2 =>a+b/c+d=a-c/b-d (đpcm)