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) 

Ta có: 

+\(\dfrac{1}{a}+\dfrac{2}{2b+1}+\dfrac{3}{3c+2}\ge2\)

\(\Rightarrow\dfrac{1}{a}\ge\dfrac{2b-1}{2b+1}+\dfrac{3c-1}{3c+2}\ge2\sqrt{\dfrac{\left(2b-1\right)\left(3c-1\right)}{\left(2b+1\right)\left(3c+2\right)}}\left(1\right)\)

+\(\dfrac{1}{a}+\dfrac{2}{2b+1}+\dfrac{3}{3c+2}\ge2\)

\(\Rightarrow\dfrac{2}{2b+1}\ge\dfrac{a-1}{a}+\dfrac{3c-1}{3c+2}\ge2\sqrt{\dfrac{\left(a-1\right)\left(3c-1\right)}{a\left(3c+2\right)}}\left(2\right)\)

+\(\dfrac{1}{a}+\dfrac{2}{2b+1}+\dfrac{3}{3c+2}\ge2\)

\(\Rightarrow\dfrac{3}{3c+2}\ge\dfrac{a-1}{a}+\dfrac{2b-1}{2b+1}\ge2\sqrt{\dfrac{\left(a-1\right)\left(2b-1\right)}{a\left(2b+1\right)}}\left(3\right)\)

Từ \(\left(1\right),\left(2\right),\left(3\right)\Rightarrow6\ge8\left(a-1\right)\left(2b-1\right)\left(3c-1\right)\)

\(\Rightarrow P=\left(a-1\right)\left(2b-1\right)\left(3c-1\right)\le\dfrac{3}{4}\)

\(\Rightarrow P_{max}=\dfrac{3}{4}\) đạt tại \(a=\dfrac{3}{2};b=1;c=\dfrac{5}{6}\)

Vì \(\dfrac{a-1}{2}=\dfrac{b-2}{3}=\dfrac{c-3}{4}\)

nên \(\dfrac{a-1}{2}=\dfrac{2b-4}{6}=\dfrac{3c-9}{12}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\dfrac{a-1}{2}=\dfrac{2b-4}{6}=\dfrac{3c-9}{12}=\dfrac{a-1-2b+4+3c-9}{2-6+12}=\dfrac{14-6}{8}=1\)

Do \(\dfrac{a-1}{2}=1\Rightarrow a=3\)

\(\dfrac{2b-4}{6}=1\Rightarrow b=5\)

\(\dfrac{3c-9}{12}=1\Rightarrow c=7\)

Vậy \(\left\{{}\begin{matrix}a=3\\b=5\\c=7\end{matrix}\right..\)

cảm ơn

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

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

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

sao lại -6=))))

`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`

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 \(\left(a;b;c\right)=\left(1;\dfrac{1}{2};\dfrac{1}{3}\right)\)

a+4/a>=2*căn a*4/a=4

b+9/b>=2*căn b*9/b=6

c+16/c>=2*căn c*16/c=8

=>3a/4+b/2+c/4+3/a+9/2b+4/c>=3+3+2=8

a+2b+3c>=20

=>a/4+b/2+3c/4>=5

=>S>=13

Dấu = xảy ra khi a=2; b=3; c=4