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) 

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

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

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

`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`

Câu a, b, c giống dạng nhau nên mình làm một câu a và câu d thôi nha, bạn tham khảo ^^

Giải:

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

Áp dụng tính chất của dãy tỉ sô bằng nhau:

\(a=\dfrac{b}{2}=\dfrac{c}{3}=\dfrac{a-b+c}{1-2+3}=\dfrac{10}{2}=5\)

\(\Rightarrow\left\{{}\begin{matrix}a=5.1=5\\b=2.5=10\\c=3.5=15\end{matrix}\right.\)

b) \(a:b:c=3:4:5\)

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

\(\Rightarrow\dfrac{a^2}{9}=\dfrac{b^2}{16}=\dfrac{c^2}{25}\)

\(\Rightarrow\dfrac{2a^2}{18}=\dfrac{2b^2}{32}=\dfrac{3c^2}{75}\)

Áp dụng tính chất của dãy tỉ sô bằng nhau:

\(\Rightarrow\dfrac{2a^2}{18}=\dfrac{2b^2}{32}=\dfrac{3c^2}{75}=\dfrac{2a^2+2b^2-3c^2}{18+32-75}=\dfrac{-100}{-25}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2=\dfrac{4.18}{2}=36\\b^2=\dfrac{4.32}{2}=64\\c^2=\dfrac{4.75}{3}=100\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\pm6\\b=\pm8\\c=\pm10\end{matrix}\right.\)

Đặt \(\left(a;2b;3c\right)=\left(x;y;z\right)\Rightarrow x+y+z=3\)

\(Q=\dfrac{x+1}{1+y^2}+\dfrac{y+1}{1+z^2}+\dfrac{z+1}{1+x^2}\)

Ta có:

\(\dfrac{x+1}{1+y^2}=x+1-\dfrac{\left(x+1\right)y^2}{1+y^2}\ge x+1-\dfrac{\left(x+1\right)y^2}{2y}=x+1-\dfrac{\left(x+1\right)y}{2}\)

Tương tự:

\(\dfrac{y+1}{1+z^2}\ge y+1-\dfrac{\left(y+1\right)z}{2}\) ; \(\dfrac{z+1}{1+x^2}\ge z+1-\dfrac{\left(z+1\right)x}{2}\)

Cộng vế:

\(Q\ge\dfrac{x+y+z}{2}+3-\dfrac{1}{2}\left(xy+yz+zx\right)\)

\(Q\ge\dfrac{x+y+z}{2}+3-\dfrac{1}{6}\left(x+y+z\right)^2=\dfrac{3}{2}+3-\dfrac{9}{6}=3\)

\(Q_{min}=3\) khi \(x=y=z=1\) hay \(\left(a;b;c\right)=\left(1;\dfrac{1}{2};\dfrac{1}{3}\right)\)