Gọi k = \(\frac{a-1}{2}=\frac{b+3}{4}=\frac{c-5}{6}\)

=> \(\begin{cases}a=2k+1\\b=4k-3\\c=6k+5\end{cases}\)

=> 5c - 4b - 3a = 30k + 25 - 16k + 12 - 6k - 3 = 8k + 34

=> 8k + 34 = 50

=> k = 2

=> \(\begin{cases}a=5\\b=5\\c=17\end{cases}\)

Ta có: \(\dfrac{2}{3}a=\dfrac{3}{4}b=\dfrac{4}{5}c\)

\(\Leftrightarrow\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{4}{3}}=\dfrac{c}{\dfrac{5}{4}}\)

mà a+b-c=38

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

\(\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{4}{3}}=\dfrac{c}{\dfrac{5}{4}}=\dfrac{a+b-c}{\dfrac{3}{2}+\dfrac{4}{3}-\dfrac{5}{4}}=\dfrac{38}{\dfrac{19}{12}}=24\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{a}{\dfrac{3}{2}}=24\\\dfrac{b}{\dfrac{4}{3}}=24\\\dfrac{c}{\dfrac{5}{4}}=24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=24\cdot\dfrac{3}{2}=36\\b=24\cdot\dfrac{4}{3}=32\\c=24\cdot\dfrac{5}{4}=30\end{matrix}\right.\)

Vậy:(a,b,c)=(36;32;30)

@Tuấn Anh Phan Nguyễn

Theo đề bài ta có:

\(\frac{a-1}{2}=\frac{b+3}{4}=\frac{c-5}{6}\)

\(\Rightarrow\frac{5c-25}{30}=\frac{3a-3}{6}=\frac{4b+12}{16} =\frac{\left(5c-25\right)-\left(3a-3\right)-\left(4b+12\right)}{30-6-16}\)

\(=\frac{\left(5c-3a-4b\right)-\left(25-3+12\right)}{8}=\frac{50-34}{8}=\frac{16}{8}=2\)

=> a = 2 . 2 + 1 = 5

b = 2 . 4 - 3 = 5

c = 2 . 6 + 5 =17

Vậy a = 5; b = 5; c = 17.

bài 2 : a) \(\dfrac{a-1}{2}=\dfrac{b+3}{4}=\dfrac{c-5}{6}\)

áp dụng dảy tỉ số bằng nhau

ta có : \(\dfrac{5\left(a-1\right)-3\left(b+3\right)-4\left(c-5\right)}{5.2-3.4-4.6}\)

\(=\dfrac{5a-5-3b-9-4c+20}{10-12-24}=\dfrac{\left(5a-3b-4c\right)-5-9+20}{-26}\)

\(=\dfrac{46+6}{-26}=\dfrac{52}{-26}=-2\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a-1}{2}=-2\\\dfrac{b+3}{4}=-2\\\dfrac{c-5}{6}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a-1=-4\\b+3=-8\\c-5=-12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=-3\\b=-11\\c=-7\end{matrix}\right.\)

vậy \(a=-3;b=-11;c=-7\)

b) ta có : \(3a=2b\Leftrightarrow6a=4b=5c\Leftrightarrow\dfrac{6a}{2}=\dfrac{4b}{2}=\dfrac{5c}{2}\)

áp dụng dảy tỉ số bằng nhau

ta có \(\dfrac{-60a-60b+60c}{-10.2-15.2+12.2}=\dfrac{60\left(-a-b+c\right)}{-20-30+24}\)

\(=\dfrac{60\left(-52\right)}{-26}=\dfrac{-3120}{-26}=120\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{6a}{2}=120\\\dfrac{4b}{2}=120\\\dfrac{5c}{2}=120\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6a=240\\4b=240\\5c=240\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=40\\b=60\\c=48\end{matrix}\right.\)

vậy \(a=40;b=60;c=48\)

\(3a=4b=5c=\frac{a}{\frac{1}{3}}=\frac{b}{\frac{1}{4}}=\frac{c}{\frac{1}{5}}\)

Áp dụng TC DTSBN ta có :

\(\frac{a}{\frac{1}{3}}=\frac{b}{\frac{1}{4}}=\frac{c}{\frac{1}{5}}=\frac{a+b+c}{\frac{1}{3}+\frac{1}{4}+\frac{1}{5}}=\frac{94}{\frac{47}{60}}=120\)

=> a = 40 ; b = 30 ; c = 24

a=21

b=37 

c=65

Ta có: 3a=2b=\(\frac{a}{2}=\frac{b}{3}\)và 4b=5c=\(\frac{b}{5}=\frac{c}{4}\)

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

\(\frac{a}{10}=4\Rightarrow a=10.4=40\)

\(\frac{b}{15}=4\Rightarrow b=15.4=60\)

\(\frac{c}{12}=4\Rightarrow c=12.4=48\)

a = 40 b = 60 c = 48

-1015+12= -13  

Có: \(3a=2b\Rightarrow\frac{a}{2}=\frac{b}{3}\Rightarrow\frac{a}{10}=\frac{b}{15}\)

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

=> \(\frac{a}{10}=\frac{b}{15}=\frac{c}{12}\)

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

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

=>\(\frac{a}{10}=4\Rightarrow a=40\)

     \(\frac{b}{15}=4\Rightarrow b=60\)

     \(\frac{c}{12}=4\Rightarrow c=48\)

ta có : \(\begin{cases}3a=2b\\4b=5c\end{cases}\)<=>\(\begin{cases}\frac{a}{2}=\frac{b}{3}\\\frac{b}{5}=\frac{c}{4}\end{cases}\)<=>\(\begin{cases}\frac{a}{10}=\frac{b}{15}\\\frac{b}{15}=\frac{c}{12}\end{cases}\)

=->\(\frac{a}{10}=\frac{b}{15}=\frac{c}{12}\)

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

=>\(\frac{a}{10}=-4\)=> a=-40

\(\frac{b}{15}=-4\)=>b=-60

\(\frac{c}{12}=-4\)=> c=-48

Ta có:

\(\frac{a}{2}=\frac{b}{3};\frac{b}{5}=\frac{c}{4}\)

\(\frac{a}{2.5}=\frac{b}{3.5};\frac{b}{5.3}=\frac{c}{4.3}\)

\(\frac{a}{10}=\frac{b}{15};\frac{b}{15}=\frac{c}{12}\Rightarrow\frac{a}{10}=\frac{b}{15}=\frac{c}{12}\Rightarrow\frac{-a}{-10}=\frac{b}{15}=\frac{c}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

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

\(\frac{-a}{-10}=4\Rightarrow-a=-40\Rightarrow a=40\)

\(\frac{b}{15}=4\Rightarrow b=60\)

\(\frac{c}{12}=4\Rightarrow c=48\)

a = 14

b = 23

c = 15

tick nha bạn

a = 14

b = 23

c = 15

tick nha

a=14

b=23

c=15

l i k e nha Phạm Minh Tuấn