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

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

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

`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`

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: \(\frac{3c-4b}{2}=\frac{4a-2c}{3}=\frac{2b-3a}{4}\)

=>\(\frac{6c-8b}{4}=\frac{12a-6c}{9}=\frac{8b-12a}{16}\)

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

\(\frac{6c-8b}{4}=\frac{12a-6c}{9}=\frac{8b-12a}{16}=\frac{6c-8b+12a-6c+8b-12a}{4+9+16}=0\)

=>12a=8b=6c

=>\(\frac{12a}{24}=\frac{8b}{24}=\frac{6c}{24}\)

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

mà c+b-a=-30

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

\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{c+b-a}{4+3-2}=\frac{-30}{5}=-6\)

=>\(\begin{cases}a=-6\cdot2=-12\\ b=-6\cdot3=-18\\ c=-6\cdot4=-24\end{cases}\)

\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b-3c}{2+6-12}=\dfrac{-20}{-4}=5\Rightarrow a=10;b=15;c=20.\)

Theo đề bài,có: \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)và \(a+2b-3c=-20\)

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

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

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

Với \(\dfrac{a}{2}=5\Rightarrow a=10\)

\(\dfrac{2b}{6}=5\Rightarrow\dfrac{b}{3}=5\Rightarrow b=15\)

\(\dfrac{3c}{12}=5\Rightarrow\dfrac{c}{4}=5\Rightarrow c=20\)

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

\(P=\dfrac{5a+10b+15c}{4}+\left(\dfrac{3}{a}+\dfrac{3a}{4}\right)+\left(\dfrac{9}{2b}+\dfrac{b}{2}\right)+\left(\dfrac{4}{c}+\dfrac{c}{4}\right)\)

\(\ge\dfrac{5\left(a+2b+3c\right)}{4}+2\sqrt{\dfrac{3}{a}.\dfrac{3a}{4}}+2\sqrt{\dfrac{9}{2b}.\dfrac{b}{2}}+2\sqrt{\dfrac{4}{c}.\dfrac{c}{4}}\)

\(\Leftrightarrow P\ge\dfrac{5.20}{4}+3+3+2=33\)

Dấu "=" xảy ra khi a=2;b=3;c=4

Vậy \(P_{min}=33\)

a>1; b>1/2; c>1/3 và \(\frac{1}{a} + \frac{2}{2b + 1} + \frac{3}{3c + 2} \ge 2 \quad (1)\)

Đặt x=a

=>x>1

=>x-1>0

Đặt y=2b+1

=>y>2*1/2+1=1+1=2

=>y-2>0

Đặt z=3c+2

=>z>3

=>z-3>0

P=(a-1)(2b-1)(3c-1)

=(x-1)(y-2)(z-3)

(1) =>\(\frac{1}{x} + \frac{2}{y} + \frac{3}{z} \ge 2 \quad (2)\)

=>\(\frac{1}{x} \ge 2 - \frac{2}{y} - \frac{3}{z} = \left(1 - \frac{2}{y}\right) + \left(1 - \frac{3}{z}\right) = \frac{y - 2}{y} + \frac{z - 3}{z}\)

=>\(\frac{1}{x} \ge \frac{y - 2}{y} + \frac{z - 3}{z} \ge 2\sqrt{\frac{(y - 2)(z - 3)}{yz}} \quad (3)\)

Chứng minh tương tự, ta sẽ có:

\(\frac{2}{y} \ge \left(1 - \frac{1}{x}\right) + \left(1 - \frac{3}{z}\right) = \frac{x - 1}{x} + \frac{z - 3}{z} \ge 2\sqrt{\frac{(x - 1)(z - 3)}{xz}} \quad (4)\)

\(\frac{3}{z} \ge \left(1 - \frac{1}{x}\right) + \left(1 - \frac{2}{y}\right) = \frac{x - 1}{x} + \frac{y - 2}{y} \ge 2\sqrt{\frac{(x - 1)(y - 2)}{xy}} \quad (5)\)

Từ (3),(4),(5) suy ra \(\frac{1}{x} \cdot \frac{2}{y} \cdot \frac{3}{z} \ge 2\sqrt{\frac{(y - 2)(z - 3)}{yz}} \cdot 2\sqrt{\frac{(x - 1)(z - 3)}{xz}} \cdot 2\sqrt{\frac{(x - 1)(y - 2)}{xy}}\)

=>\(\frac{6}{xyz}\ge8\cdot\frac{(x - 1)(y - 2)(z - 3)}{xyz}\)

=>\(6\ge8(x-1)(y-2)(z-3)\)

=>\((x-1)(y-2)(z-3)\le\frac{6}{8}=\frac{3}{4}\)

=>P<=3/4

Dấu '=' xảy ra khi \(\begin{cases} \dfrac{x - 1}{x} = \dfrac{y - 2}{y} = \dfrac{z - 3}{z} \\ \dfrac{1}{x} + \dfrac{2}{y} + \dfrac{3}{z} = 2 \end{cases}\)

Đặt \(\dfrac{x - 1}{x}=\dfrac{y - 2}{y}=\dfrac{z - 3}{z}=k\)

=>\(\dfrac{1}{x}=\dfrac{2}{y}=\dfrac{3}{z}=1-k\)

\(\frac{1}{x}+\frac{2}{y}+\frac{3}{z}=2\)

=>1-k+1-k+1-k=2

=>3-3k=2

=>3k=1

=>k=1/3

=>\(\frac{1}{x}=\frac{2}{y}=\frac{3}{z}=1-\frac13=\frac23\)

=>x=3/2 và y=3 và z=9/2

=>a=3/2; b=1; c=5/6