Cách 1:Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Rồi thay vào hai vế mà chứng minh

Cách 2:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)

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

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

a) áp dụng tính chất của dãy tỉ số bằng nhau ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-c^2}{b^2-d^2}\)

Do \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\)=> đpcm 

b)  áp dụng tính chất của dãy tỉ số bằng nhau ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\left(\frac{a-c}{b-d}\right)^2\)=> đpcm

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

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

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

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

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

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

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

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

Còn với\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\) thì bạn chỉ cần thay dấu trừ thành dấu công là được

Chúc bạn học tốt

a) Ta có: \(\frac{a}{b}=\frac{c}{d}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

+)\(\frac{2a^2-3b^2}{2c^2-3d^2}=\frac{2.\left(bk\right)^2-3b^2}{2.\left(dk\right)^2-3d^2}=\frac{2.b^2.k^2-3.b^2}{2.d^2.k^2-3.d^2}\)

                                                                \(=\frac{2.b^2.\left(k^2-3\right)}{2.d^2.\left(k^2-3\right)}\)

                                                                  \(=\frac{b^2}{d^2}\)(1)

+)\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}\)(2)

Từ (1) và (2), ta có: \(\frac{2a^2-3b^2}{2c^2-3d^2}=\frac{ab}{cd}\)

Học tốt nha!!!

a/b=c/d

=>a/c=b/d=a+b/c+d

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

=>ab/cd=(a+b)^2/(c+d)^2  

Vay......

a/b=c/d

=> a/c=b/d=a+b/c+d

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

=> ab/cd=(a+b)^2/(c+d)^2

# Hok_tốt nha

a) ta có: \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow\frac{a}{b}=k\Rightarrow a=bk\)

\(\frac{c}{d}=k\Rightarrow c=dk\)

thay vào   \(\frac{a^2-b^2}{ab}=\frac{\left(bk^2\right)-b^2}{bkb}=\frac{bkbk-bb}{bkb}=\frac{bb\times\left(kk-1\right)}{bbk}=\frac{kk-1}{k}\)

                   \(\frac{c^2-d^2}{cd}=\frac{\left(dk^2\right)-d^2}{dkd}=\frac{dkdk-dd}{dkd}=\frac{dd\times\left(kk-1\right)}{ddk}=\frac{kk-1}{k}\)

\(\Rightarrow\frac{a^2-b^2}{ab}=\frac{c^2-d^2}{cd}\left(=\frac{kk-1}{k}\right)\)

b) ta có \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow\frac{a}{b}=k\Rightarrow a=bk\)

\(\Rightarrow\frac{c}{d}=k\Rightarrow c=dk\)

thay vào  \(\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left(bk+b\right)^2}{bkbk+bb}=\frac{b\left(k+1\right)\times b\left(k+1\right)}{bb\left(kk+1\right)}=\frac{bb\left(k+1\right)\left(k+1\right)}{bb\left(kk+1\right)}=\frac{\left(k+1\right)\left(k+1\right)}{kk+1}\)

     \(\frac{\left(c+d\right)^2}{c^2+d^2}=\frac{\left(dk+d\right)^2}{dkdk+dd}=\frac{\left(d\left(k+1\right)\right)^2}{dd\left(kk+1\right)}=\frac{d\left(k+1\right)\times d\left(k+1\right)}{dd\left(kk+1\right)}=\frac{dd\left(k+1\right)\left(k+1\right)}{dd\left(kk+1\right)}=\frac{\left(k+1\right)\left(k+1\right)}{kk+1}\)

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

(a² + b²) / (c² + d²) = ab/cd 
<=> (a² + b²)cd = ab(c² + d²) 
<=> a²cd + b²cd = abc² + abd² 
<=> a²cd - abc² - abd² + b²cd = 0 
<=> ac(ad - bc) - bd(ad - bc) = 0 
<=> (ac - bd)(ad - bc) = 0 
<=> ac - bd = 0 hoặc ad - bc = 0 
<=> ac = bd hoặc ad = bc 
<=> a/b = d/c hoặc a/b = c/d (đpcm)

Câu a:

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