Question

tìm a , b , c,biết 3a = 2b ; 4b = 5c và -a - b + c = -52

Answer 1

3a = 2b = > 6a = 4b ; 4b = 5c

=> 6a = 4b = 5c

=> 6a/60 = 4b/60 = 5c/60

=> a/10 = b/15 = c/12

=> -a/-10 = b/15 = c/12 

=> (-a - b + c)/(-10 - 15 + 12) = a/10 = b/15 = c/12

=> -52/-13 = a/10 = b/15 = c/12

=> 4 = a/10 = b/15 = c/12

=> x = 40; b = 60; c = 48

Answer 2

\(3a=2b;4b=5c\)và \(-a-b+c=-52\)

Theo bài ra ta cs 

\(+,3a=2b\Rightarrow\frac{a}{2}=\frac{b}{3}\)

\(+,4b=5c\Rightarrow\frac{b}{5}=\frac{c}{4}\)

Ta lại cs : 

\(+,\frac{a}{2}=\frac{b}{3}\Rightarrow\frac{a}{10}=\frac{b}{15}\)(1)

\(+,\frac{b}{5}=\frac{c}{4}\Rightarrow\frac{b}{15}=\frac{c}{12}\)(2)

Từ (1) và (2) \(\Rightarrow\frac{a}{10}=\frac{b}{15}=\frac{c}{12}\)

ADTC dãy tỉ số bằng nhau ta cs 

\(\frac{a}{10}=\frac{b}{15}=\frac{c}{12}=\frac{-a-b+c}{-10-15+12}=-\frac{52}{-13}=4\)

\(\Rightarrow\hept{\begin{cases}\frac{a}{10}=4\\\frac{b}{15}=4\\\frac{c}{12}=4\end{cases}\Rightarrow\hept{\begin{cases}a=4.10=40\\b=4.15=60\\c=4.12=48\end{cases}}}\)