\(\frac{a}{b}=\frac{c}{d}\)(\(b,d\ne0\))

\(\Leftrightarrow ad=bc\)

\(\Leftrightarrow2ad=2bc\)

\(\Leftrightarrow ad-bc=bc-ad\)

\(\Leftrightarrow ad-bc+ac-bd=bc-ad+ac-bd\)

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

\(\Leftrightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\)(\(a-b,c-d\ne0\))

Đề bài thiếu rồi bạn ơi

ta có a/b=c/d nên ad=bc ( tính chất nhân chéo của phân số)

a/b=c/d

a/c=b/d

d/c=b/a

d/b=c/a

Vì sao mk chịu

\(=>A+B-C+D=a+b-5-b-c+1-b+c+4+b-a\)

\(=-5+4=-1\)

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

d/b=c/a

a/b=d/c

a/c=b/d

a/d=b/c

d/b=c/a

 help me!!!

còn lâu 

hahaha!!!

đúng thì k nha

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)nên \(a=bk;c=dk\)

a) Ta có: \(\frac{a}{a-b}=\frac{bk}{bk-b}=\frac{bk}{b\left(k-1\right)}=\frac{k}{k-1}\left(1\right)\)

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

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

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

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

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

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

Giải:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

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

Ta có:

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

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

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

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

theo đề bài ta có
\(ab\left(c^2+d^2\right)=ab.c^2+ab.d^2=\left(a.c\right).\left(b.c\right)+\left(a.d\right).\left(b.d\right)\\ cd\left(a^2+b^2\right)=cd.a^2+cd.b^2=\left(c.a\right).\left(d.a\right)+\left(c.b\right).\left(d.b\right)\)
\(\left(a.c\right)\left(b.c\right)+\left(a.d\right)\left(b.d\right)=\left(c.a\right)\left(d.a\right)+\left(c.b\right)\left(d.b\right)\) vì mỗi vế đều bằng nhau
- Cnứng minh \(\frac{\left(a^2+b^2\right)}{c^2+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
ta có vì \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{\left(a+b\right)}{\left(c+d\right)}=\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2}{c^2}=\frac{b^2}{d^2}\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(a^2+b^2\right)}{\left(c^2+d^2\right)}\)

Gọi a/b=c/d=k(k khác 0)

Ta có:

a=bk

c=dk

VT:(\(\frac{a+b}{c+d}\))² =(\(\frac{bk+b}{dk+d}\))² =(\(\frac{b\left(k+1\right)}{d\left(k+1\right)}\))² =(\(\frac{b}{d}\))^{2 (1)}

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

Từ (1) và (2) suy ra bằng nhau

cho tỉ lệ thức ab = cd 

chứng minh rằng (2008a+2009c)(b+d)=(a+c)(2008+2009d)

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

Ta có :

\(\frac{2008a+2009c}{a+c}=\frac{2008bk+2009dk}{bk+dk}=\frac{k\left(2008b+2009d\right)}{k\left(b+d\right)}=\frac{2008b+2009d}{b+d}\)

\(\Rightarrow\frac{2008a+2009c}{a+c}=\frac{2008b+2009d}{b+d}\Rightarrow\left(2008a+2009c\right)\left(b+d\right)=\left(a+c\right)\left(2008b+2009d\right)\)

=> ĐPCM

Ta có : \(\frac{a}{b}=\frac{c}{d}=\frac{2008a}{2008b}=\frac{2009c}{2009d}\)

\(\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{2008a}{2008b}=\frac{2009c}{2009d}=\frac{a+c}{b+d}=\frac{2008a+2009c}{2008b+2009d}\)

\(\Rightarrow\left(2008a+2009c\right).\left(b+d\right)=\left(a+c\right).\left(2008b+2009d\right)\)