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

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

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

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

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

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

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

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

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

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

Ta có: \(A=\frac{2}{1\cdot5}+\frac{3}{5\cdot11}+\frac{4}{11\cdot19}+\frac{5}{19\cdot29}+\frac{6}{29\cdot41}\)

\(=\frac12\left(\frac{4}{1\cdot5}+\frac{6}{5\cdot11}+\frac{8}{11\cdot19}+\frac{10}{19\cdot29}+\frac{12}{29\cdot41}\right)\)

\(=\frac12\left(1-\frac15+\frac15-\frac{1}{11}+\frac{1}{11}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{41}\right)\)

\(=\frac12\left(1-\frac{1}{41}\right)=\frac12\cdot\frac{40}{41}=\frac{20}{41}\)

Ta có: \(B=\frac{40}{31\cdot39}+\frac{35}{39\cdot46}+\frac{30}{46\cdot52}+\frac{25}{52\cdot57}+\frac{20}{57\cdot61}\)

\(=5\left(\frac{8}{31\cdot39}+\frac{7}{39\cdot46}+\frac{6}{46\cdot52}+\frac{5}{52\cdot57}+\frac{4}{57\cdot61}\right)\)

\(=5\left(\frac{1}{31}-\frac{1}{39}+\frac{1}{39}-\frac{1}{46}+\frac{1}{46}-\frac{1}{52}+\frac{1}{52}-\frac{1}{57}+\frac{1}{57}-\frac{1}{61}\right)\)

\(=5\left(\frac{1}{31}-\frac{1}{61}\right)=5\cdot\frac{30}{61\cdot31}=\frac{150}{1891}\)

\(\frac{A}{B}=\frac{20}{41}:\frac{150}{1891}=\frac{20}{41}\cdot\frac{1891}{150}=\frac{2}{15}\cdot\frac{1891}{41}=\frac{3782}{615}\)

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 ... 

Tỉ số giữa a và c là:

\(\frac27\cdot\frac23=\frac{4}{21}\)

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

_c/m ... a,b,c nha