\(\text{Ta có}:\)

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

\(\text{Theo tính chất của dãy tỉ số bằng nhau ta có:}\)

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

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

\(\text{Câu b sai đề nha bạn ơi. Đề đúng phải là (a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)}\)

Theo đề ra ta có:

\(\frac{a+b+c+d}{a-b+c-d}=\frac{a+b-c-d}{a-b-c+d}=\frac{a+b+c+d+a+b-c-d}{a-b+c-d+a-b-c+d}=\frac{2a+2d}{2a-2b}=\frac{a+b}{a-b}\left(1\right)\)

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

Từ (1) và (2) ta suy ra được :\(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)tương tự câu a ta suy ra \(\frac{a}{b}=\frac{c}{d}\)

cảm ơn bạn nha

a)\(\frac{ab}{cd}=\frac{bk.b}{dk.b}=\frac{b^2}{d^2}\left(1\right)\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\Rightarrow\left(a+2c\right)\left(b-d\right)=\left(a-c\right)\left(b+2d\right)\left(đpcm\right)\)

Chúc bạn học tốt !!! 

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

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

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

\(tu\left(1\right)\left(2\right)suyra\)\(\frac{2a+b}{a-2b}=\frac{2c+d}{c-2d}\)

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

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

\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b+a-b}{c+d+c-d}=\frac{2a}{2c}=\frac{a}{c}\left(1\right)\)

\(\text{Chứng minh tương tự: }\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b-\left(a-b\right)}{c+d-\left(c-d\right)}=\frac{a+b-a+b}{c+d-c+d}=\frac{2b}{2d}=\frac{b}{d}\left(2\right)\)

\(\text{Từ (1) và (2): }\Rightarrow\frac{a}{b}=\frac{c}{d}\left(Đ\text{PCM}\right)\)

b) \(\left(a+2c\right)\left(b-d\right)=\left(a-c\right)\left(b+2d\right)\)

\(\Rightarrow ab+ad+2cd=ab+2da+cd+2dc\)

\(\Rightarrow ad+2cb=2da+cb\)

\(\Rightarrow ab=cd\)

Ta có:

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

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

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

Từ (1) , (2) \(\Rightarrow\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)

b) \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\left(1\right)\)

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

Từ (1) , (2) \(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

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

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

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

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

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

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

các câu còn lại bạn tự làm đi! HI.......