\(Ta có: \(\frac{1}{2a+3b+3c}=\frac{1}{\left(a+b\right)+\left(a+c\right)+2\left(b+c\right)}\) Theo Cauchy: \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\) => \(\frac{1}{2a+3b+3c}\le\frac{1}{4}\left(\frac{1}{\left(a+b\right)+\left(a+c\right)}+\frac{1}{2\left(b+c\right)}\right)\le\frac{1}{4}\left(\frac{1} {4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)+\frac{1}{2\left(b+c\right)}\right)\) => \(\frac{1}{2a+3b+3c}\le\frac{1}{8}\left(\frac{1}{2\left(a+b\right)}+\frac{1}{2\left(a+c\right)}+\frac{1}{b+c}\right)\) Tương tự: \(\frac{1}{3a+2b+3c}\le\frac{1}{8}\left(\frac{1}{2\left(a+b\right)}+\frac{1}{2\left(b+c\right)}+\frac{1}{a+c}\right)\) Và: \(\frac{1}{3a+3b+2c}\le\frac{1}{8}\left(\frac{1}{2\left(a+c\right)}+\frac{1}{2\left(b+c\right)}+\frac{1}{a+b}\right)\) => \(P\le\frac{1}{8}\left(\frac{2}{a+b}+\frac{2}{a+c}+\frac{2}{b+c}\right)=\frac{1}{4}.2017\) => Pmax  = 2017:4=504,25\)

Ta có: \(\frac{1}{2a+3b+3c}=\frac{1}{\left(a+b\right)+\left(a+c\right)+2\left(b+c\right)}\)

Theo Cauchy: \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)

=> \(\frac{1}{2a+3b+3c}\le\frac{1}{4}\left(\frac{1}{\left(a+b\right)+\left(a+c\right)}+\frac{1}{2\left(b+c\right)}\right)\le\frac{1}{4}\left(\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)+\frac{1}{2\left(b+c\right)}\right)\)

=> \(\frac{1}{2a+3b+3c}\le\frac{1}{8}\left(\frac{1}{2\left(a+b\right)}+\frac{1}{2\left(a+c\right)}+\frac{1}{b+c}\right)\)

Tương tự: \(\frac{1}{3a+2b+3c}\le\frac{1}{8}\left(\frac{1}{2\left(a+b\right)}+\frac{1}{2\left(b+c\right)}+\frac{1}{a+c}\right)\)

Và: \(\frac{1}{3a+3b+2c}\le\frac{1}{8}\left(\frac{1}{2\left(a+c\right)}+\frac{1}{2\left(b+c\right)}+\frac{1}{a+b}\right)\)

=> \(P\le\frac{1}{8}\left(\frac{2}{a+b}+\frac{2}{a+c}+\frac{2}{b+c}\right)=\frac{1}{4}.2017\)

=> P_max = 2017:4=504,25

viết hẳn ra p/s đi bạn khó hiểu quá

+4 hoặc -4

Vì 2a/3b=3b/4c=4c/5d=5d/2a nên suy ra 2a=3b=4c=5d ( Theo công thức dãy tỉ số bằng nhau)

=> 2a/3b=3b/4c=4c/5d=5d/2a=1

=>C=1+1+1+1=4

Vậy C=4

Ta có

\(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=\frac{2a+3b+4c+5d}{3b+4c+5d+2a}=1.\) (Tính chất dãy tỷ số bằng nhau)

\(\Rightarrow\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}=4.1=4\)

\(A=\left(-2a+3b-4c\right)-\left(-2a-3b-4c\right)\)

\(=-2a+3b-4c+2a+3b+4c\)

\(=6b\)

b) Khi \(a=2012,b=-1,c=-2013\) ta có :

\(A=6b=6\cdot\left(-1\right)=-6\)

Vậy \(A=-6\) khi \(a=2012,b=-1,c=-2013\)

Giải:

a) \(A=\left(-2a+3b-4c\right)-\left(-2a-3b-4c\right)\) 

\(A=-2a+3b-4c+2a+3b+4c\) 

\(A=\left(-2a+2a\right)+\left(3b+3b\right)+\left(-4c+4c\right)\)

 \(A=0+2.3b+0\) 

\(A=6b\)

b) Ta thay: \(a=2012;b=-1;c=-2013\) 

Ta có:

\(A=\left(-2a+3b-4c\right)-\left(-2a-3b-4c\right)\) 

\(A=\left(-2.2012+-3.1--4.2013\right)-\left(-2.2012--3.1--4.2013\right)\) 

\(A=\left(-2.2012-3.1+4.2013\right)-\left(-2.2012+3.1+4.2013\right)\)

\(A=-2.2012-3.1+4.2013+2.2012-3.1-4.2013\) 

\(A=\left(-2.2012+2.2012\right)+\left(-3.1-3.1\right)+\left(4.2013-4.2013\right)\) 

\(A=0+2.-3.1+0\) 

\(A=-6\)

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

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

=>2a+b=3c; 2b+c=3a; 2c+a=3b

\(\left(\frac{2a+b}{c}\right)+\left(\frac{a}{2b+c}\right)+\left(\frac{3b}{2c+a}\right)\)

\(=\frac{3c}{c}+\frac{a}{3a}+\frac{3b}{3b}=3+\frac13+1=4+\frac13=\frac{13}{3}\)