thự hiện phép tính (a-b)-(a-b)+(2a-b)-(2a-3b)
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12 tháng 4 2018
\(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\)
BT
11 tháng 4 2018
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
TV
1
PP
3
LH
25 tháng 8 2016
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
NP
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
22 tháng 6 2016
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)
\(\Rightarrow\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}=4.1=4\)
11 tháng 6 2021
\(A=\left(-2a+3b-4c\right)-\left(-2a-3b-4c\right)\)
\(=-2a+3b-4c+2a+3b+4c\)
\(=6b\)
b) Khi \(a=2012,b=-1,c=-2013\) ta có :
\(A=6b=6\cdot\left(-1\right)=-6\)
Vậy \(A=-6\) khi \(a=2012,b=-1,c=-2013\)
11 tháng 6 2021
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:
\(A=\left(-2a+3b-4c\right)-\left(-2a-3b-4c\right)\)
\(A=\left(-2.2012+-3.1--4.2013\right)-\left(-2.2012--3.1--4.2013\right)\)
\(A=\left(-2.2012-3.1+4.2013\right)-\left(-2.2012+3.1+4.2013\right)\)
\(A=-2.2012-3.1+4.2013+2.2012-3.1-4.2013\)
\(A=\left(-2.2012+2.2012\right)+\left(-3.1-3.1\right)+\left(4.2013-4.2013\right)\)
\(A=0+2.-3.1+0\)
\(A=-6\)
LM
0
(a - b) - (a - b) + (2a - b) - (2a - 3b)
= a - b - a + b + 2a - b - 2a + 3b
= (a - a + 2a - 2a) - (b - b + b - 3b)
= - (-2b) = 2b