c, \(\dfrac{2a}{3}=\dfrac{3b}{4}=\dfrac{4c}{5}\&a+b+c=49\)
=> \(\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{4}{3}}=\dfrac{c}{\dfrac{5}{4}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{4}{3}}=\dfrac{c}{\dfrac{5}{4}}=\dfrac{a+b+c}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)
=> a = 12 . \(\dfrac{3}{2}=18\)
b = 12 . \(\dfrac{4}{3}=16\)
c = 12 . \(\dfrac{5}{4}=15\)
Vậy ...............
d, \(\dfrac{a}{b}=\dfrac{9}{7}\&\dfrac{b}{c}=\dfrac{7}{3},a-b+c=15\)
Ta có : \(\dfrac{a}{b}=\dfrac{9}{7}\Rightarrow\dfrac{a}{9}=\dfrac{b}{7}\)
\(\dfrac{b}{c}=\dfrac{7}{3}\Rightarrow\dfrac{b}{7}=\dfrac{c}{3}\)
=> \(\dfrac{a}{9}=\dfrac{b}{7}=\dfrac{c}{3}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{9}=\dfrac{b}{7}=\dfrac{c}{3}=\dfrac{a-b+c}{9-7+3}=\dfrac{15}{5}=3\)
=> \(\dfrac{a}{9}=3\Rightarrow a=27\)
\(\dfrac{b}{7}=3\Rightarrow b=21\)
\(\dfrac{c}{3}=3\Rightarrow c=9\)
Vậy..............