\(2a=3b\Rightarrow\dfrac{a}{3}=\dfrac{b}{2}\Rightarrow\dfrac{a}{21}=\dfrac{b}{14}\)

\(5b=7c\Rightarrow\dfrac{b}{7}=\dfrac{c}{5}\Rightarrow\dfrac{b}{14}=\dfrac{c}{10}\)

Áp dụng t/c dtsbn:

\(\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}=\dfrac{3a}{63}=\dfrac{7b}{98}=\dfrac{5c}{50}=\dfrac{3a+5c-7b}{63+50-98}=\dfrac{30}{15}=2\)

\(\Rightarrow\left\{{}\begin{matrix}a=21.2=42\\b=14.2=28\\c=10.2=20\end{matrix}\right.\)

Ta có : 4a = 3b => 28a = 21b (1)

            7b = 5c => 21b = 15c (2)

Từ (1) và (2) => 28a = 21b = 15c 

Ta có : 28a = 21b = 15c \(=\frac{a}{\frac{1}{28}}=\frac{b}{\frac{1}{21}}=\frac{c}{\frac{1}{15}}=\frac{2a}{\frac{1}{14}}=\frac{3b}{\frac{1}{7}}=\frac{2a+3b-c}{\frac{1}{14}+\frac{1}{7}-\frac{1}{15}}=\frac{186}{\frac{31}{210}}=1260\)

Nên : 28a = 1260 => a = 45

         21b = 1260 => b = 60

         15c = 1260 => c = 84

Vậy ........................

Ta có:

 \(4a=3b\)=> \(\frac{a}{3}=\frac{b}{4}\)=> \(\frac{a}{15}=\frac{b}{20}\left(1\right)\)

\(7b=5c\)=>\(\frac{b}{5}=\frac{c}{7}\) => \(\frac{b}{20}=\frac{c}{28}\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\)

=>\(\frac{a}{15}=\frac{b}{20}=\frac{c}{28}\)=>\(\frac{2a}{30}=\frac{3b}{60}=\frac{c}{28}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{2a}{30}=\frac{3b}{60}=\frac{c}{28}=\frac{2a+3b-c}{30+60-28}=\frac{186}{62}=3\)

=>\(\frac{a}{15}=3\)=>\(a=45\)

    \(\frac{b}{20}=3\)=>\(b=60\)

    \(\frac{c}{28}=3\)=>\(c=84\)

Vậy \(a=40;b=60;c=84\)

Ta có: \(2a=3b\)=> \(\frac{a}{3}=\frac{b}{2}\)=>\(\frac{a}{21}=\frac{b}{14}\left(1\right)\)

          \(5b=7c\)=>\(\frac{b}{7}=\frac{c}{5}\) =>\(\frac{b}{14}=\frac{c}{10}\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\)

=>\(\frac{a}{21}=\frac{b}{14}=\frac{c}{10}\)=> \(\frac{3a}{63}=\frac{7b}{98}=\frac{5c}{50}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{3a}{63}=\frac{7b}{98}=\frac{5c}{50}=\frac{3a-7b+5c}{63-98+50}=\frac{30}{15}=2\)

=>\(\frac{a}{21}=2\)=>\(a=42\)

    \(\frac{b}{14}=2\)=>\(b=28\)

    \(\frac{c}{10}=2\)=>\(c=20\)

Vậy \(a=42;b=28;c=20\)

3a+5c=7b+30

=>3a+5c-7b=30

\(2a=3b=>\frac{a}{3}=\frac{b}{2}=>\frac{a}{3}.\frac{1}{7}=\frac{b}{2}.\frac{1}{7}=>\frac{a}{21}=\frac{b}{14}\)

\(5b=7c=>\frac{b}{7}=\frac{c}{5}=>\frac{b}{7}.\frac{1}{2}=\frac{c}{5}.\frac{1}{2}=>\frac{b}{14}=\frac{c}{10}\)

\(=>\frac{a}{21}=\frac{b}{14}=\frac{c}{10}=\frac{3a}{63}=\frac{7b}{98}=\frac{5c}{50}=\frac{3a+5c-7b}{63+50-98}=\frac{30}{15}=2\)

\(=>\frac{a}{21}=2=>a=21.2=42\)

\(=>\frac{b}{14}=2=>b=14.2=28\)

\(=>\frac{c}{10}=2=>c=10.2=20\)

Vậy a=42,b=28,c=20.

minh tran

ta có 2a=3b =>a=3b/2 
5b=7c =>c=5b/7 
=>3.3b/2+5.5b/7+7b=30 
=>9b/2+25b/7+7b=30 
=>63b/14+50b/14+93b/14=30 
=>211b/14=30 
=>211/14.b=30 
=>211/14.30=b 
=>6330/14=b 
=>3165/7=b 
=>9495/7=3b=2a 
=>a=9495/14 
tương tự c= vượt giới hạn tính

a=9495/14

c= k tinh dc

2a = 3b => a/3 = b/2 => a/21 = b/14

5b = 7c => b/7 = c/5 => b/14 = c/10

=> a/21 = b/14 = c/10 => 3a/63 = 7b/98 = 5c/50

Áp dung TCDTSBN ta có:

\(\frac{3a}{63}=\frac{7b}{98}=\frac{5c}{50}=\frac{3a-7b+5c}{63-98+50}=\frac{30}{15}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{a}{21}=2\\\frac{b}{14}=2\\\frac{c}{10}=2\end{cases}}\Rightarrow\hept{\begin{cases}a=42\\b=28\\c=20\end{cases}}\)

Vậy....

giúp

giúp mình

Ta có: 2a=3b

nên \(\dfrac{a}{3}=\dfrac{b}{2}\)

hay \(\dfrac{a}{21}=\dfrac{b}{14}\left(1\right)\)

Ta có: 5b=7c

nên \(\dfrac{b}{7}=\dfrac{c}{5}\)

hay \(\dfrac{b}{14}=\dfrac{c}{10}\left(2\right)\)

Từ \(\left(1\right),\left(2\right)\) suy ra \(\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}\)

hay \(\dfrac{3a}{63}=\dfrac{7b}{98}=\dfrac{5c}{50}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{3a}{63}=\dfrac{7b}{98}=\dfrac{5c}{50}=\dfrac{3a-7b+5c}{63-98+50}=\dfrac{30}{15}=2\)

Do đó: a=42; b=28; c=20

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