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

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

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

`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`

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) 

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

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

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

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

\(\text{Theo bài ra ta có : }\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\\ \Rightarrow\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}\\ a+2b-3c=-20\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=5\\\dfrac{2b}{6}=5\\\dfrac{3c}{12}=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=10\\\dfrac{b}{3}=5\\\dfrac{c}{4}=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=10\\b=15\\c=20\end{matrix}\right.\)

Vậy \(a=10;b=15;c=20\)

\(\)

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

Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\Rightarrow a=2k;b=3k;c=4k\)

Thay a+2b-3c=-20 bằng k

\(\Rightarrow2k+2\cdot3k-3\cdot4k=-20\)

\(\Rightarrow2k+6k-12k=-20\)

\(\Rightarrow k\left(2+6-12\right)=-20\)

\(\Rightarrow k\cdot\left(-4\right)=-20\)

\(\Rightarrow k=5\)

Từ đó suy ra:

*a=2k\(\Rightarrow a=10\)

*b=3k\(\Rightarrow b=15\)

*c=4k\(\Rightarrow c=20\)

Vậy a=10;b=15;c=20

chúch bạn hoc tốt

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{a+2b-3c}{2+6+12}=\dfrac{-20}{20}=-1\)

( Vì a + 2b - 3c = -20 )

Do đó :

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

\(\dfrac{b}{3}=-1\Rightarrow b=-3\)

\(\dfrac{c}{4}=-1\Rightarrow c=-4\)

Vậy ....

Tính lại :

-2 + 2(-3) - 3(-4) = -20

Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=2k\\b=3k\\c=4k\end{matrix}\right.\)

Ta có: 2k + 2.3k - 3.4k = -20

2k + 6k -12k =-20

-4k = -20

k = 5 (1)

Thay (1) vào, ta có:

\(\left\{{}\begin{matrix}a=2.5\\b=3.5\\c=4.5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=10\\b=15\\c=20\end{matrix}\right.\)

Vậy ...

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

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

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

Dấu "=" xảy ra khi a=2;b=3;c=4

Vậy \(P_{min}=33\)