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\(2a=4b\Rightarrow\frac{a}{10}=\frac{b}{5}\)
\(3b=5c\Rightarrow\frac{b}{5}=\frac{c}{3}\)
\(\Leftrightarrow\frac{a}{10}=\frac{b}{5}=\frac{c}{3}=\frac{a+2b-3c}{10+2.5-3.3}=\frac{99}{11}=9\)
a=90
b=45
c=27
Giải:
Ta có: \(2a=4b\Rightarrow a=2b\Rightarrow\frac{a}{1}=\frac{b}{2}\Rightarrow\frac{a}{5}=\frac{b}{10}\)
\(3b=5c\Rightarrow\frac{b}{5}=\frac{c}{3}\Rightarrow\frac{b}{10}=\frac{c}{6}\)
\(\Rightarrow\frac{a}{5}=\frac{b}{10}=\frac{c}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{a}{5}=\frac{b}{10}=\frac{c}{6}=\frac{2b}{20}=\frac{3c}{18}=\frac{a+2b-3c}{5+20-18}=\frac{99}{7}\)
+) \(\frac{a}{5}=\frac{99}{7}\Rightarrow a=\frac{495}{7}\)
+) \(\frac{b}{10}=\frac{99}{7}\Rightarrow b=\frac{990}{7}\)
+) \(\frac{c}{6}=\frac{99}{7}\Rightarrow c=\frac{594}{7}\)
Vậy bộ số \(\left(a;b;c\right)\) là \(\left(\frac{495}{7};\frac{990}{7};\frac{594}{7}\right)\)
Có 2a=4b => \(\dfrac{a}{4}=\dfrac{b}{2}\)=> \(\dfrac{a}{4.5}=\dfrac{b}{2.5}\)=>\(\dfrac{a}{20}=\dfrac{b}{10}\) (1)
Có 3b=5c => \(\dfrac{b}{5}=\dfrac{c}{3}\)=>\(\dfrac{b}{5.2}=\dfrac{c}{3.2}\)=>\(\dfrac{b}{10}=\dfrac{c}{6}\) (2)
Từ (1) và (2) => \(\dfrac{a}{20}=\dfrac{b}{10}=\dfrac{c}{6}\)
Đặt \(\dfrac{a}{20}=\dfrac{b}{10}=\dfrac{c}{6}\) = k
=> a=20k , b=10k, c=6k
Thay a=20k , b=10k, c=6k vào a+2b-3c=99, ta có :
20k+2.10k-3.6k=99
=> 20k+20k-18k=99
=> k(20+20-18)=99
=> k= 99:22=4,5
=>a=20.4,5=90, b=10.4,5=45, c=6.4,5=27
Vậy a=90, b=45, c=27
a )\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=\frac{2a}{2c}\)
\(\frac{a-b}{c-d}=\frac{2a}{2c}\Rightarrow\frac{a-b}{2a}=\frac{c-d}{2c}\) ( đpcm)
b ) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{3a}{3c}=\frac{2b}{2d}=\frac{5a-3b}{5c-3d}=\frac{3a+2b}{3c+2d}\)
\(\Rightarrow\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}\) ( đpcm )
\(2a=4b\Rightarrow\dfrac{a}{4}=\dfrac{b}{2}\Rightarrow\dfrac{a}{20}=\dfrac{b}{10};3b=5c\Rightarrow\dfrac{b}{5}=\dfrac{c}{3}\Rightarrow\dfrac{b}{10}=\dfrac{c}{6}\\ \Rightarrow\dfrac{a}{20}=\dfrac{b}{10}=\dfrac{c}{6}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{20}=\dfrac{b}{10}=\dfrac{c}{6}=\dfrac{2b}{20}=\dfrac{3c}{18}=\dfrac{a+2b-3c}{20+20-18}=\dfrac{99}{22}=\dfrac{9}{2}\\ \Rightarrow\left\{{}\begin{matrix}a=90\\b=45\\c=27\end{matrix}\right.\)