\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}.\frac{c}{d}=\frac{a}{b}.\frac{a}{b}=\frac{a^2}{b^2};\frac{a}{b}.\frac{c}{d}=\frac{c}{d}.\frac{c}{d}=\frac{c^2}{d^2}\\ \Rightarrow\frac{a}{b}.\frac{c}{d}=\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)

Ta có:

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

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

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

\(\Rightarrow\frac{ca+cb}{ca+ad}=\frac{bc+bd}{ad+bd}=\frac{ca+bd}{ca-bd}=1\)

\(\Rightarrow ca+cb=ca+ad\)

\(\Rightarrow cb=ad\)

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

Ta có tỉ lệ thức 

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

Suy ra 

a=bk

c=dk

Nên ta có

\(\frac{a.b}{c.d}=\frac{bk.b}{dk.d}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}\)

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

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

Vì a/b=c/d =>a/c=b/d => a^2/c^2=a/c.b/d=a.b/c.d (1)

Ta có:

       a/c=b/d => a^2/c^2=b^2/d^2 = a^2-b^2/c^2-d^2 (2) ( Áp dụng tc dãy tỉ số bằng nhau)

   Từ (1) và (2) => a.b/c.d=a^2-b^2/c^2-d^2

Bài 1: 
Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{k}{3k+1}\)

\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{k}{3k+1}\)

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

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

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

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

a) Giải:

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

Ta có:
\(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2}{d^2}\) (1)

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

Từ (1) và (2) suy ra \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\left(đpcm\right)\)

b) Giải:
Để \(P\in Z\Rightarrow2x-3⋮x+1\)

Ta có:
\(2x-3⋮x+1\)

\(\Rightarrow\left(2x+2\right)-5⋮x+1\)

\(\Rightarrow5⋮x+1\)

\(\Rightarrow x+1\in\left\{1;-1;5;-5\right\}\)

+) \(x+1=1\Rightarrow x=0\)

+) \(x+1=-1\Rightarrow x=-2\)

+) \(x+1=5\Rightarrow x=4\)

+) \(x+1=-5\Rightarrow x=-6\)

Vậy \(x\in\left\{0;-2;4;-6\right\}\)

\(\Rightarrow5⋮x+1\)

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

\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)(tính chất dãy tỉ số bằng nhau)

\(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\left(đpcm\right)\)

2)\(P=\frac{2x-3}{x+1}=\frac{2x+2-5}{x+1}=\frac{2\left(x+1\right)-5}{x+1}=2-\frac{5}{x+1}\)

\(\Rightarrow P\in Z\Leftrightarrow2-\frac{5}{x+1}\in Z\Leftrightarrow\frac{5}{x+1}\in Z\Leftrightarrow5⋮x+1\Leftrightarrow x+1\inƯ\left(5\right)\)

\(\Rightarrow x+1\in\left\{-1;-5;1;5\right\}\)

\(\Rightarrow x\in\left\{-2;-6;0;4\right\}\)