23 = 3b ; 5b = 7c ; và 3a + 5c + 7b = 50
=> Ta được:
\(\dfrac{a}{3}\)= \(\dfrac{b}{2}\); \(\dfrac{b}{7}\)= \(\dfrac{c}{5}\)
=> \(\dfrac{a}{21}\)= \(\dfrac{b}{14}\); \(\dfrac{b}{14}\)= \(\dfrac{c}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{21}\)+\(\dfrac{b}{14}\)+\(\dfrac{c}{10}\)= \(\dfrac{\left(3a+5b+7c\right)}{\left(3.21+5.14+7.10\right)}\)=\(\dfrac{50}{203}\)
=> \(\dfrac{a}{21}\)= \(\dfrac{50}{203}\)=> a = \(\dfrac{150}{29}\)
\(\dfrac{b}{14}\)= \(\dfrac{50}{203}\)=> b = \(\dfrac{100}{29}\)
\(\dfrac{c}{10}\) = \(\dfrac{50}{203}\)=> c = \(\dfrac{500}{203}\)
Vậy a = \(\dfrac{150}{29}\)
b = \(\dfrac{100}{29}\)
c = \(\dfrac{500}{203}\)
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