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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 = \(1-1-1-1=-2\)

a) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\frac{3a}{3c}=\frac{4b}{4d}=\frac{3a+4b}{3c+4d}=\frac{3a-4b}{3c-4d}.\)

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

b) ta có: \(\frac{a}{b}=\frac{c}{d}=\frac{5a}{5b}=\frac{2c}{2d}=\frac{4a}{4b}\)

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

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

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

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

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

a)  đặt \(\frac{a}{b}=\frac{c}{d}=k\Leftrightarrow a=b.k;c=d.k\)

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

    b)          bó tay

a/ Đặt :

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

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có :

\(VT=\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\left(1\right)\)

\(VP=\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

b/ Đặt :

\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có :

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

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

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)