\(2a=3b\Rightarrow\dfrac{a}{3}=\dfrac{b}{2}\Rightarrow\dfrac{3a}{9}=\dfrac{7b}{14}\Rightarrow\dfrac{3a}{63}=\dfrac{7b}{98}\left(1\right)\)
\(5b=7c\Rightarrow\dfrac{b}{7}=\dfrac{c}{5}\Rightarrow\dfrac{7b}{49}=\dfrac{5c}{25}\Rightarrow\dfrac{7b}{98}=\dfrac{5c}{50}\left(2\right)\)
Từ (1)(2) \(\Rightarrow\dfrac{3a}{63}=\dfrac{7b}{98}=\dfrac{5c}{50}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ,ta có :
\(\dfrac{3a}{63}=\dfrac{7b}{98}=\dfrac{5c}{50}=\dfrac{3a+5c-7b}{63+50-98}=\dfrac{30}{15}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3a}{63}=2\Rightarrow3a=126\Rightarrow a=42\\\dfrac{7b}{98}=2\Rightarrow7b=196\Rightarrow b=28\\\dfrac{5c}{50}=2\Rightarrow5c=100\Rightarrow c=20\end{matrix}\right.\)
Vậy \(a=42;b=28;c=20\)
Theo bài ra ta có:
\(2a=3b\Rightarrow\dfrac{a}{3}=\dfrac{b}{2}\Rightarrow\dfrac{a}{21}=\dfrac{b}{14}\)
\(5b=7c\Rightarrow\dfrac{b}{7}=\dfrac{c}{5}\Rightarrow\dfrac{b}{14}=\dfrac{c}{10}\)
\(\Leftrightarrow\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}=\dfrac{3a+5c-7b}{3.21+5.10-7.14}=\dfrac{30}{15}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{21}=2\Rightarrow a=42\\\dfrac{b}{14}=2\Rightarrow b=28\\\dfrac{c}{10}=2\Rightarrow c=20\end{matrix}\right.\)
Vậy ..........................
Chúc bạn học tốt!
Từ 2a = 3b => \(\dfrac{a}{3}\) = \(\dfrac{b}{2}\) (1)
Từ 5b = 7c => \(\dfrac{b}{7}\) = \(\dfrac{c}{5}\) (2)
Từ (1) và (2) => \(\dfrac{a}{21}\)=\(\dfrac{b}{14}\)=\(\dfrac{c}{10}\) = \(\dfrac{3a}{63}\)=\(\dfrac{7b}{98}\)=\(\dfrac{5c}{50}\)= \(\dfrac{3a+7b-5c}{63+98-50}\)=\(\dfrac{30}{15}\) = 2 (Áp dụng dãy t/s bằng nhau nha )
=> a = 21.2 = 42
=> b = 14.2 = 28
=> c = 10.2 = 20
Vậy a= 42 , b= 28 , c= 20 .
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