\(\dfrac{a}{b}=2\Rightarrow\dfrac{a}{2}=b\Rightarrow\dfrac{a}{6}=\dfrac{b}{3}\)

\(\dfrac{b}{c}=3\Rightarrow\dfrac{b}{3}=c\)

\(\Rightarrow\dfrac{a}{6}=\dfrac{b}{3}=c=k\\ \Rightarrow a=6k;b=3k;c=k\)

\(\dfrac{a+b}{b+c}=\dfrac{6k+3k}{3k+k}=\dfrac{9k}{4k}=\dfrac{9}{4}\)

Đặt \(\left\{{}\begin{matrix}\dfrac{a}{b^2}=x\\\dfrac{b}{c^2}=y\\\dfrac{c}{a^2}=z\end{matrix}\right.\Rightarrow xyz=1;x+y+z=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)

Ta có \(x+y+z=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)

\(\Leftrightarrow x+y+z=xy+yz+zx\)

\(\Leftrightarrow xyz-1+x+y+z-xy-yz-zx=0\)

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)=0\)

​\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)=0\)​

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=1\\z=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b^2}=1\\\dfrac{b}{c^2}=1\\\dfrac{c}{a^2}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=b^2\\b=c^2\\c=a^2\end{matrix}\right.\left(đpcm\right)\)

Theo bài ra ta có \(\dfrac{a}{b}=\dfrac{2}{7};\dfrac{b}{c}=\dfrac{2}{3}\Rightarrow\dfrac{a}{2}=\dfrac{b}{7};\dfrac{b}{2}=\dfrac{c}{3}\Rightarrow\dfrac{a}{4}=\dfrac{b}{14}=\dfrac{c}{21}\)

\(\Rightarrow\dfrac{a}{4}=\dfrac{c}{21}\Leftrightarrow\dfrac{a}{c}=\dfrac{4}{21}\)

Vậy ... 

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

\(\dfrac{a}{15}=\dfrac{b}{10}=\dfrac{c}{8}=\dfrac{a+b+c}{15+10+8}=\dfrac{11}{33}=\dfrac{1}{3}\)

Do đó: a=5; b=10/3; c=8/3

`a)a/b<c/d`
Nhân 2 vế cho `bd>0` ta có:
`(abd)/b<(bcd)/d`
`<=>ad<bc`
`b)ad<bc`
Chia 2 vế cho `bd>0` ta có:
`(ad)/(bd)<(bc)/(bd)`
`<=>a/b<c/d`.