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Đề có vấn đề nha em !!!

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

=> A:B=0

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 \(\frac{a-b}{b-c}=\frac{25}{8}\)

=25/8 á bạn
chắc luôn

Từ \(a:b=9:4\Rightarrow\dfrac{a}{b}=\dfrac{9}{4}\Rightarrow\dfrac{a}{9}=\dfrac{b}{4}\Rightarrow\dfrac{a}{45}=\dfrac{b}{20}\)

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

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

Đặt \(\dfrac{a}{45}=\dfrac{b}{20}=\dfrac{c}{12}=k\Rightarrow a=45k;b=20k;c=12k\)

Khi đó \(A=\dfrac{a-b}{b-c}=\dfrac{45k-20k}{20k-12k}=\dfrac{25k}{8k}=\dfrac{25}{8}\)

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

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

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

Thay \(\left\{{}\begin{matrix}a=45k\\b=20k\\c=12k\end{matrix}\right.\) vào \(\dfrac{a-b}{b-c}\)

Ta được: \(\dfrac{45k-20k}{20k-12k}=\dfrac{\left(45-20\right)k}{\left(20-12\right)k}=\dfrac{25k}{8k}=\dfrac{25}{8}\)

Vậy: .........................................

\(\frac{a}{b}=\frac{9}{4}\Rightarrow\frac{a}{9}=\frac{b}{4}\Rightarrow\frac{a}{45}=\frac{b}{20}\)(1)

\(\frac{b}{c}=\frac{5}{3}\Rightarrow\frac{b}{5}=\frac{c}{3}\Rightarrow\frac{b}{20}=\frac{c}{12}\)(2)

Từ (1) và (2) => \(\frac{a}{45}=\frac{b}{20}=\frac{c}{12}\)

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

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

Ta có \(\dfrac{a}{b}=\dfrac{9}{4}\)=>\(\dfrac{a}{9}=\dfrac{b}{4}\)=>\(\dfrac{a}{45}=\dfrac{b}{20}\)(1)

\(\dfrac{b}{c}=\dfrac{5}{3}\)=>\(\dfrac{b}{5}=\dfrac{c}{3}\) =>\(\dfrac{b}{20}=\dfrac{c}{12}\)(2)

Từ (1) và (2) ta có :

\(\dfrac{a}{45}=\dfrac{b}{20}=\dfrac{c}{12}\)( Quy đồng mẫu)

Đặt \(\dfrac{a}{45}=\dfrac{b}{20}=\dfrac{c}{12}\)=k

=> a=45k , b=20k , c=12k (*)

Thay (*) vào \(\dfrac{a-b}{b-c}\) ta có :

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

Vậy tỉ số của \(\dfrac{a-b}{b-c}\) là \(\dfrac{25}{8}\)