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\(\frac{a}{b}=\frac{9}{4}\Rightarrow\frac{a}{9}=\frac{b}{4}\Rightarrow\frac{a}{45}=\frac{b}{20}\)(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 :
\(\frac{a-b}{b-c}=\frac{45k-20k}{20k-12k}=\frac{k\left(45-20\right)}{k\left(20-12\right)}=\frac{45-20}{20-12}=\frac{25}{8}\)
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 \(\frac{a-b}{b-c}=\frac{25}{8}\)
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 \(\frac{a-b}{b-c}=\frac{25}{8}\)
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: \(\dfrac{a-b}{b-c}=\dfrac{25}{8}\)
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 \(\dfrac{25}{8}\)
a) ĐKXĐ: \(x\ne-1\)
Ta có:
\(\frac{x+1}{8}=\frac{8}{x+1}\)
\(\Rightarrow\left(x+1\right)^2=8^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=8\\x+1=-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=-9\end{cases}\left(TMĐKXĐ\right)}\)
\(\)
a, \(\frac{x+1}{8}=\frac{8}{x+1}\)
\(\Leftrightarrow\left(x+1\right)^2=8.8\)
\(\Leftrightarrow\left(x+1\right)=\pm8\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=8\\x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-9\end{cases}}}\)
b, \(\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\left(2x+3y=186\right)\)
Theo đề bài ta có:
\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{3.5}=\frac{y}{4.5}\Rightarrow\frac{x}{15}=\frac{y}{20}\)
\(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{5.4}=\frac{z}{7.4}\Rightarrow\frac{y}{20}=\frac{z}{28}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{15}=\frac{y}{20}=\frac{2x}{30}=\frac{3y}{60}=\frac{2x+3y}{90}=\frac{186}{90}=\frac{31}{15}\)
\(\Rightarrow\frac{2x}{30}=\frac{31}{15}\Rightarrow2x=62\Rightarrow x=31\)
\(\frac{3y}{60}=\frac{31}{15}\Rightarrow3y=124\Rightarrow y=\frac{124}{3}\)
Mà \(\frac{y}{20}=\frac{z}{28}\Rightarrow\frac{\frac{124}{3}}{20}=\frac{z}{28}\Rightarrow\frac{31}{15}=\frac{z}{28}\)
Từ đây bạn tìm nốt z nha
\(\frac{a}{5}=\frac{b}{3},\frac{b}{7}=\frac{c}{9}\Rightarrow\frac{a}{35}=\frac{b}{21},\frac{b}{21}=\frac{c}{27}\Rightarrow\frac{a}{35}=\frac{b}{21}=\frac{c}{27}\)
\(\Rightarrow\frac{a}{35}=\frac{b}{21}=\frac{c}{27}=\frac{a+b}{35+21}=\frac{a+b}{56}=\frac{b-c}{21-27}=\frac{b-c}{-6}\)(T/C)
\(\Rightarrow\frac{a+b}{56}=\frac{b-c}{-6}=\frac{a+b}{b-c}=\frac{56}{-6}=-\frac{28}{3}\)
Giải:
Ta có: \(a:b=5:3\Rightarrow\frac{a}{5}=\frac{b}{3}\Rightarrow\frac{a}{35}=\frac{b}{21}\)
\(b:c=7:9\Rightarrow\frac{b}{7}=\frac{c}{9}\Rightarrow\frac{b}{21}=\frac{c}{27}\)
\(\Rightarrow\frac{a}{35}=\frac{b}{21}=\frac{c}{27}\)
Đặt \(\frac{a}{35}=\frac{b}{21}=\frac{c}{27}=k\)
\(\Rightarrow a=35k,b=21k,c=27k\)
Từ đó \(\frac{a+b}{b-c}=\frac{35k+21k}{21k-27k}=\frac{56k}{-6k}=\frac{-28}{3}\)
Vậy \(\frac{a+b}{b-c}=\frac{-28}{3}\)