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

\(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}=\dfrac{a-b-c}{8-12-15}=\dfrac{28}{-19}=\dfrac{-28}{19}\)

Do đó: \(\left\{{}\begin{matrix}a=\dfrac{-224}{19}\\b=\dfrac{-336}{19}\\c=\dfrac{-420}{19}\end{matrix}\right.\)

Ta có:\(\dfrac{x^2}{4}=\dfrac{x}{2};\dfrac{y^2}{9}=\dfrac{y}{3};\dfrac{z^2}{25}=\dfrac{z}{5}\)

Aps dụng tính chất dãy tỉ số bằn nhau:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-y+z}{2-3+5}=\dfrac{4}{4}=1\)

=>\(\dfrac{x}{2}=1=>x=2\)

  \(\dfrac{y}{3}=1=>y=3\)

\(\dfrac{z}{5}=1=>z=5\)

Vậy x=2, y=3, z=5

Ta có : \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{25}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-y+z}{2-3+5}=\dfrac{4}{4}=1\)

\(\Leftrightarrow x=2;y=3;z=5\)

tham khảo!!

https://lazi.vn/edu/exercise/tim-cac-so-a-b-c-biet-rang-a-2-b-3-c-4-va-a-2-b-2-2c-2-108

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{a+2b-3c}{2+2\cdot3-3\cdot4}=\dfrac{-20}{-4}=5\\ \Rightarrow\left\{{}\begin{matrix}a=10\\b=15\\c=20\end{matrix}\right.\)

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b-3c}{2+6-12}=\dfrac{-20}{-4}=5\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=5\\\dfrac{b}{3}=5\\\dfrac{c}{4}=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=10\\b=15\\c=20\end{matrix}\right.\)

`a/2 = b/3 = c/4`

`=> a/2 = (2b)/6 = (3c)/12`

mà `a+2b-3c=-20`

áp dụng tính chất dãy tỉ số bằng nhau ta có

` a/2 = (2b)/6 = (3c)/12 = (a+2b-3c)/(2+6-12)=(-20)/-4 = 5`

` => a=5xx2=10`

`b=5xx3=15`

`c=5xx4=20`

ta có : `a/2=b/3=c/4 =>a/2=(2b)/6=(3c)/12` và `a+2b-3c=-20`

ADTC dãy tỉ số bằng nhau ta có :

`a/2=(2b)/6=(3c)/6=(a+2b-3c)/(2+6-12)=(-20)/-4=5`

`=>a/2=5=>a=5.2=10`

`=>b/3=5=>b=5.3=15`

`=>c/4=5=>c=5.4=20`

#\(N\)

`a/2 = b/3 = c/4 , a+2b-3c = -20`

`-> a/2 =`\(\dfrac{2b}{6}=\dfrac{3c}{12}\)

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

\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b-3c}{2+6-12}=\dfrac{-20}{-4}=5\)

`-> a/2 = 5 , b/3 = 5 , c/4 = 5`

`-> a=2.5 = 10 , b=3.5=15 , c=4.5=20`

a) \(\dfrac{a}{5}=\dfrac{b}{4}\Rightarrow\dfrac{a^2}{25}=\dfrac{b^2}{16}\)

Áp dụng tính chất DTSBN :

\(\dfrac{a^2}{25}=\dfrac{b^2}{16}=\dfrac{a^2-b^2}{25-16}=\dfrac{1}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{1}{9}\cdot25=\dfrac{25}{9}\\b^2=\dfrac{1}{9}\cdot16=\dfrac{16}{9}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\dfrac{5}{3};b=\dfrac{4}{3}\\a=\dfrac{-5}{3};b=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy \(\left(a;b\right)\in\left\{\left(\dfrac{5}{3};\dfrac{4}{3}\right);\left(-\dfrac{5}{3};-\dfrac{4}{3}\right)\right\}\)

b) \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\)

Áp dụng tính chất DTSBN :

\(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}=\dfrac{2c^2}{32}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4\)

\(\Rightarrow\left\{{}\begin{matrix}a^2=4.4=16\\b^2=4.9=36\\c^2=4,16=64\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=4;=6;c=8\\a=-4;b=-6;c=-8\end{matrix}\right.\)

Vậy (a;b;c) \(\in\left\{\left(4;6;8\right);\left(-4;-6;-8\right)\right\}\)

Ta có: \(\dfrac{a}{3}=\dfrac{b}{4}\)

          \(\dfrac{b}{2}=\dfrac{c}{5}\Rightarrow\dfrac{b}{4}=\dfrac{c}{10}\)

\(\Rightarrow\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{10}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{10}=\dfrac{a-c+b}{3-10+4}=\dfrac{3}{-3}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}a=\left(-1\right).3=-3\\b=\left(-1\right).4=-4\\c=\left(-1\right).10=-10\end{matrix}\right.\)

\(\dfrac{a}{3}=\dfrac{b}{4};\dfrac{b}{2}=\dfrac{c}{5}\Rightarrow\dfrac{b}{4}=\dfrac{c}{10}\\ \Rightarrow\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{10}\)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{10}=\dfrac{a+b-c}{3+4-10}=\dfrac{3}{-3}=-1\\ \Rightarrow\left\{{}\begin{matrix}a=-1\cdot3=-3\\b=-1\cdot4=-4\\c=-1\cdot10=-10\end{matrix}\right.\)

\(\dfrac{b}{2}=\dfrac{c}{5}\Rightarrow\dfrac{b}{4}=\dfrac{c}{10}\)

\(\Rightarrow\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{10}=\dfrac{a-c+b}{3-10+4}=\dfrac{3}{-3}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}a=\left(-1\right).3=-3\\b=\left(-1\right).4=-4\\c=\left(-1\right).10=-10\end{matrix}\right.\)

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

\(\Leftrightarrow\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{4}{3}}=\dfrac{c}{\dfrac{5}{4}}=\dfrac{a+b+c}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

Khi đó \(a=12.\dfrac{3}{2}=18;b=12.\dfrac{4}{3}=16;c=12.\dfrac{5}{4}=15\)

Vậy (a,b,c) = (18,16,15)