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

\(\frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\)

\(\frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d\left(2k+5\right)}{d\left(3k-4\right)}=\frac{2k+5}{3k-4}\)

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

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

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

\(\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{5b}{5d}=\frac{2a+5b}{2c+5d}\)

\(\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{4b}{4d}=\frac{3a-4b}{3c-4d}\)

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

\(\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\left(đpcm\right)\)

nên đặt (1) , (2) 

rồi từ (1) và (2) =>.....

như vậy khó hiểu lắm

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

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

b, \(\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{5b}{5d}=\frac{2a+5b}{2c+5d}\) 

\(\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{4b}{4d}=\frac{3a-4b}{3c-4d}\)

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

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

Đặt \(\frac{a}{b}=\frac{c}{d}=v\)

\(\Rightarrow\hept{\begin{cases}a=vb\\c=vd\end{cases}}\)( 1 )

Thay (1) vào vế trái , ta có :

\(VT=\frac{2vb+5b}{3vb-4b}=\frac{b\left(2v+5\right)}{b\left(3v-4\right)}=\frac{2v+5}{3v-4}\)( *)

Thay (1) vào vế phải ta có :

\(VP=\frac{2vd+5d}{3vd-4d}=\frac{2v+5}{3v-4}\)(**)

Từ (  * ) và (** )

=> ĐPCM

Các bạn giúp mình nhé ! Mình đang cần gấp

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

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

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

Mà a+b+c+d khác 0

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

=> b = a = c = d

Ta có:

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

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

\(P=\frac{7a}{7a}-\frac{7b}{7b}-\frac{7c}{7c}-\frac{7d}{7d}\)

\(P=1-1-1-1=-2\)

Ta có:

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

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

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

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

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

Khi đó : P = \(\frac{2a+5a}{3a+4a}-\frac{2a+5c}{3a+4a}-\frac{2a+5a}{3a+4a}-\frac{2a+5a}{3a+4a}\)

P = \(1-1-1-1=-2\)

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

Khi đó: \(\frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\)

            \(\frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d\left(2k+5\right)}{d\left(3k-4\right)}=\frac{2k+5}{3k-4}\)

\(\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\left(=\frac{2k+5}{3k-4}\right)\)

Từ\(\frac{a}{b}\)=\(\frac{c}{d}\)suy ra \(\frac{a}{c}\)=\(\frac{b}{d}\)( t/c TLT)

Áp dụng tính chất của dãy TSBN ta có:\(\frac{a}{c}\)=\(\frac{b}{d}\)=\(\frac{2a+5b}{2c+5d}\)=\(\frac{3a-4b}{3c-4d}\)

Từ \(\frac{2a+5b}{2c+5d}\)=\(\frac{3a-4b}{3c-4d}\) suy ra\(\frac{2a+5b}{3a-4b}\)=\(\frac{2c+5d}{3c-4d}\)(t/c TLT)

Giả sử tất cả các tỷ lệ thức đều có nghĩa.

a)

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

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

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

k có giả thuyết thì sao chứng minh 

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow\begin{cases}a=bk\\c=dk\end{cases}\)\(\Rightarrow\frac{2bk+5b}{3bk-4b}=\frac{2dk+5d}{3dk-4d}\)

Xét VT \(\frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\left(1\right)\)

Xét VP \(\frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d\left(2k+5\right)}{d\left(3k-4\right)}=\frac{2k+5}{3k-4}\left(2\right)\)

Từ (1) và (2) ta có Đpcm

ở lớp 7 hình như đang học cm tam giác đồng dạng,nên áp dụng cái này(nếu bạn chưa học thì có thể hỏi cô):

\(\frac{a}{b}=\frac{x}{y}=>\frac{a}{a+-b}=\frac{x}{x+-y}\)

\(\frac{a}{b}=\frac{c}{d}=>\frac{3a}{4b}=\frac{3c}{4d}=>\frac{3a}{3a-4b}=\frac{3c}{3c-4d}\)\(=>\frac{2a}{3a-4b}=\frac{2c}{3c-4d}\)(1)

Làm tương tự ;\(\frac{5b}{3a-4b}=\frac{5d}{3c-4d}\)(2)  Cộng(1)(2) thì được