Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

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

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

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

b: \(\frac{3a+7b}{5a-7b}=\frac{3\cdot bk+7b}{5\cdot bk-7b}=\frac{b\left(3k+7\right)}{b\left(5k-7\right)}=\frac{3k+7}{5k-7}\)

\(\frac{3c+7d}{5c-7d}=\frac{3\cdot dk+7d}{5\cdot dk-7d}=\frac{d\left(3k+7\right)}{d\left(5k-7\right)}=\frac{3k+7}{5k-7}\)

Do đó: \(\frac{3a+7b}{5a-7b}=\frac{3c+7d}{5c-7d}\)

d: \(\frac{4a+9b}{4a-7b}=\frac{4\cdot bk+9b}{4\cdot bk-7b}=\frac{b\left(4k+9\right)}{b\left(4k-7\right)}=\frac{4k+9}{4k-7}\)

\(\frac{4c+9d}{4c-7d}=\frac{4\cdot dk+9d}{4\cdot dk-7d}=\frac{d\left(4k+9\right)}{d\left(4k-7\right)}=\frac{4k+9}{4k-7}\)

Do đó: \(\frac{4a+9b}{4a-7b}=\frac{4c+9d}{4c-7d}\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)

  suy ra:   x/5 = 45   =>  x  =  225

               y/7 = 45  =>  y  =  315

               z/9 = 45  =>  z  =  405

Áp dụng t/c dãy tỉ số bằng nhau ta được:

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

\(\Rightarrow\hept{\begin{cases}2a=b+c\\2b=c+a\\2c=a+b\end{cases}}\Leftrightarrow\hept{\begin{cases}3a=a+b+c\\3b=a+b+c\\3c=a+b+c\end{cases}}\Rightarrow a=b=c\)

Thay vào ta được:

\(Q=\frac{a+2b}{c}+\frac{b+2c}{a}+\frac{c+2a}{b}=\frac{3c}{c}+\frac{3a}{a}+\frac{3b}{b}=9\)

\(M=\frac{a+5b}{c}+\frac{b+5c}{a}+\frac{c+5a}{b}=\frac{6c}{c}+\frac{6a}{a}+\frac{6b}{b}=18\)

Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)

\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)

\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)

Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)

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

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

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

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