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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 \(\dfrac{25}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\frac{a}{b}=\frac{9}{4}\Rightarrow\frac{a}{9}=\frac{b}{4}\Rightarrow\frac{a}{45}=\frac{b}{20}\) (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) \(\Rightarrow\frac{a}{45}=\frac{b}{20}=\frac{c}{12}\)
Đặt \(\frac{a}{45}=\frac{b}{20}=\frac{c}{12}=k\Rightarrow a=45k;b=20k;c=12k\) Thay vào \(\frac{a-b}{b-c}\) ta được :
\(\frac{a-b}{b-c}=\frac{45k-20k}{20k-12k}=\frac{25k}{8k}=\frac{25}{8}\)
Vậy \(\frac{a-b}{b-c}=\frac{25}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a}{b}=\frac{9}{4}\Rightarrow\frac{a}{9}=\frac{b}{4}\Rightarrow\frac{a}{45}=\frac{b}{20}\)(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 :
\(\frac{a-b}{b-c}=\frac{45k-20k}{20k-12k}=\frac{k\left(45-20\right)}{k\left(20-12\right)}=\frac{45-20}{20-12}=\frac{25}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\frac{a-b}{b-c}=\frac{25}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a:b=2:5; b:c=4:3=>\(\frac{a}{2}=\frac{b}{5};\frac{b}{4}=\frac{c}{3}\Rightarrow\frac{a}{8}=\frac{b}{20}=\frac{c}{15}\)
Đặt \(k=\frac{a}{8}=\frac{b}{20}=\frac{c}{15}\Rightarrow k^2=\frac{a.b}{8.20}=\frac{c^2}{225}\)
Áp dụng tính chất dãy tỉ số bằng nhau,ta có:
\(k^2=\frac{a.b}{160}=\frac{c^2}{225}=\frac{a.b-c^2}{160-225}=\frac{-10,4}{-65}=0,16\)
\(\Rightarrow\left[\begin{array}{nghiempt}k=0,4\\k=-0,4\end{array}\right.\)
Với k=0,4=>a=3,2; b=8; c=6=>|a+b+c|=17,2
Với k=-0,4 =>a=-3,2; b=-8; c=-6=>|a+b+c|=17,2
Vậy|a+b+c|=17,2
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\frac{a-b}{b-c}=\frac{25}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
Ta có: \(\frac{a}{9}=\frac{b}{4}\Rightarrow\frac{a}{45}=\frac{b}{20}\)
\(\frac{b}{5}=\frac{c}{3}\Rightarrow\frac{b}{20}=\frac{c}{12}\)
\(\Rightarrow\frac{a}{45}=\frac{b}{20}=\frac{c}{12}\)
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: \(\frac{a-b}{b-c}=\frac{45k-20k}{20k-12k}=\frac{25k}{8k}=\frac{25}{8}\)
Vậy \(\frac{a-b}{b-c}=\frac{25}{8}\)
Theo bài ra:
\(\dfrac{a}{b}=\dfrac{9}{4}\Rightarrow a=\dfrac{9}{4}.b\)
\(\dfrac{b}{c}=\dfrac{5}{3}\Rightarrow c=b:\dfrac{5}{3}\)
Thay \(a=\dfrac{9}{4b};c=b:\dfrac{5}{3}\) vào \(\dfrac{a-b}{b-c}\), ta có:
\(\dfrac{\dfrac{9b}{4}-b}{b-\dfrac{3b}{5}}=\dfrac{\dfrac{9b}{4}-\dfrac{4b}{4}}{\dfrac{5b}{5}-\dfrac{3b}{5}}=\dfrac{5b}{4}:\dfrac{2b}{5}=\dfrac{5b}{4}.\dfrac{5}{2b}=\dfrac{25}{8}\)
Vậy: \(\dfrac{a-b}{b-c}=\dfrac{25}{8}\)