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\(A=4^{15}.9^9-8.3^{18}.8^9=2^{2.15}.3^{2.9}-2^3.3^{18}.2^{3.9}=2^{30}.3^{18}-2^{30}.3^{18}=0\)

=> A:B=0

- Xét: a : b = 9 : 4 \(\Rightarrow\frac{a}{9}=\frac{b}{4}\)\(\Rightarrow\frac{a}{45}=\frac{b}{20}\)

       b : c = 5 : 3 \(\Rightarrow\frac{b}{5}=\frac{c}{3}\)\(\Rightarrow\frac{b}{20}=\frac{c}{12}\)    

=> \(\frac{a}{45}=\frac{b}{20}=\frac{c}{12}\)

- Đặt: \(\frac{a}{45}=\frac{b}{20}=\frac{c}{12}=k\Rightarrow\hept{\begin{cases}a=45.k\\b=20.k\\c=12.k\end{cases}}\)

-Thay a = 45.k, b = 20.k , c = 12.k vào \(\frac{a-b}{b-c}\) ;ta có: 

\(\frac{a-b}{b-c}=\frac{45.k-20.k}{20.k-12.k}=\frac{25.k}{8.k}=\frac{25}{8}\)

Vậy \(\frac{a-b}{b-c}=\frac{25}{8}\)

\(A=\)\(\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}\)

\(A=\left(\frac{9}{1}-8\right)+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)\)

\(A=1+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)

\(A=\frac{10}{10}+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)

\(A=10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}\right)>\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}=B\)

\(\Rightarrow A:B=\frac{10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)\(=10\)

Vậy \(A:B=10\)

dat \(\frac{a}{3}\)=\(\frac{b}{4}\)=k =>a=3k va b=4k

ma \(\frac{b}{8}\)=\(\frac{c}{9}\) nen \(\frac{4k}{8}\)=\(\frac{c}{9}\)=> c=\(\frac{9k}{2}\)

theo bai ra c+a=60 =>3k+\(\frac{9k}{2}\)=60 =>\(\frac{6k+9k}{2}\)=60 =>15k=120 => k= 8

nen a=3*8=24   b=4*8=32       c=\(\frac{9\cdot8}{2}\)=36

( A - B ) / 3 = 9

A - B = 9 * 3 = 27

Từ đây lập hiệu tỉ là ra

Giải:
Ta có: \(a:b=9:4\Rightarrow\frac{a}{9}=\frac{b}{4}\Rightarrow\frac{a}{45}=\frac{b}{20}\)

\(b:c=5:3\Rightarrow\frac{b}{5}=\frac{c}{3}\Rightarrow\frac{b}{20}=\frac{c}{12}\)

\(\Rightarrow\frac{a}{45}=\frac{b}{20}=\frac{c}{12}\)

Đặt \(\frac{a}{45}=\frac{b}{20}=\frac{c}{12}=k\Rightarrow\left\{\begin{matrix}a=45k\\b=20k\\c=12k\end{matrix}\right.\)

Lại có: \(\frac{a-b}{b-c}=\frac{45k-20k}{20k-12k}=\frac{\left(45-20\right)k}{\left(20-12\right)k}=\frac{25}{8}\)

Vậy \(\frac{a-b}{b-c}=\frac{25}{8}\)

Giải:
Ta có: \(a:b=9:4\Rightarrow\frac{a}{9}=\frac{b}{4}\Rightarrow\frac{a}{45}=\frac{b}{20}\)

\(b:c=5:3\Rightarrow\frac{b}{5}=\frac{c}{3}\Rightarrow\frac{b}{20}=\frac{c}{12}\)

\(\Rightarrow\frac{a}{45}=\frac{b}{20}=\frac{c}{12}\)

Đặt \(\frac{a}{45}=\frac{b}{20}=\frac{c}{12}=k\Rightarrow a=45k,b=20k,c=12k\)

\(\frac{a-b}{b-c}=\frac{45k-20k}{20k-12k}=\frac{\left(45-20\right)k}{\left(20-12\right)k}=\frac{25}{8}\)

Vậy \(\frac{a-b}{b-c}=\frac{25}{8}\)