Cho các số thực a,b,c thõa mãn a+b+c=2018 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{2017}{2018}\)
Tính giá trị của biểu thức P=\(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{b+a}\)
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MA
25 tháng 9 2017
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2018}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\left(a+b+c=2018\right)\)
\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{a+b+c-c}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left[\dfrac{1}{ab}+\dfrac{1}{c\left(a+b+c\right)}\right]\left(a+b\right)=0\)
\(\Leftrightarrow\dfrac{ac+bc+c^2+ab}{abc\left(a+b+c\right)}\times\left(a+b\right)=0\)
\(\Leftrightarrow\dfrac{\left(a+c\right)\left(b+c\right)\left(a+b\right)}{abc\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-c\\b=-c\\a=-b\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}b=2018\\a=2018\\c=2018\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{1}{2018^{2017}}\)
14 tháng 4 2018
hình như bạn bị sai rồi
a=-c
a=-b
b=-c
=>a=-b=-(-c)=c
mà a=-c =>vô lý
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 11 2021
\(1-\dfrac{1}{1+a}\ge\dfrac{2017}{b+2017}+\dfrac{2018}{c+2018}\ge2\sqrt{\dfrac{2017.2018}{\left(b+2017\right)\left(c+2018\right)}}\)
\(1-\dfrac{2017}{b+2017}\ge\dfrac{1}{1+a}+\dfrac{2018}{b+2018}\ge2\sqrt{\dfrac{2018}{\left(1+a\right)\left(b+2018\right)}}\)
\(1-\dfrac{2018}{c+2018}\ge\dfrac{1}{1+a}+\dfrac{2017}{b+2017}\ge2\sqrt{\dfrac{2017}{\left(1+a\right)\left(b+2017\right)}}\)
Nhân vế:
\(\dfrac{abc}{\left(a+1\right)\left(b+2017\right)\left(c+2018\right)}\ge\dfrac{8.2017.2018}{\left(a+1\right)\left(b+2017\right)\left(c+2018\right)}\)
\(\Rightarrow abc\ge8.2017.2018\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 11 2021
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(2.1;2.2017;2.2018\right)=...\)
6 tháng 2 2022
Xét \(\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=126.16=2016\)
\(\Leftrightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=2016\)
\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=2013\)
Vậy A = 2013
B
15 tháng 5 2021
Bài làm :
Ta có :
\(\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\frac{4}{a+b}\le\frac{1}{a}+\frac{1}{b}\)
\(\Leftrightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\left(1\right)\)
Dấu "=" xảy ra khi : a=b
Chứng minh tương tự như trên ; ta có :
\(\hept{\begin{cases}\frac{1}{b+c}\text{≤}\frac{1}{4}\left(\frac{1}{b}+\frac{1}{c}\right)\left(2\right)\\\frac{1}{c+a}\text{≤}\frac{1}{4}\left(\frac{1}{c}+\frac{1}{a}\right)\left(3\right)\end{cases}}\)
Cộng vế với vế của (1) ; (2) ; (3) ; ta được :
\(A\text{≤}\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\text{=}\frac{3}{2}\)
Dấu "=" xảy ra khi ;
\(\hept{\begin{cases}a=b=c\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\end{cases}}\Leftrightarrow a=b=c=1\)
Vậy Max (A) = 3/2 khi a=b=c=1
AH
Akai Haruma
Giáo viên
13 tháng 5 2023
Thỏa mãn $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ hay $a+b+c=1$ vậy bạn?
4 tháng 12 2021
Áp dụng t/c dtsbn ta có:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)
\(\dfrac{a+b-c}{c}=1\Rightarrow a+b-c=c\Rightarrow a+b=2c\\ \dfrac{b+c-a}{a}=1\Rightarrow b+c-a=a\Rightarrow b+c=2a\\ \dfrac{c+a-b}{b}=1\Rightarrow c+a-b=b\Rightarrow c+a=2b\)
\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)\\ =\dfrac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc}\\ =\dfrac{2c.2b.2a}{abc}\\ =\dfrac{8abc}{abc}\\ =8\)
Vì a + b + c = 2018
\(\Rightarrow\left\{{}\begin{matrix}b+c=2018-a\\c+a=2018-b\\a+b=2018-c\end{matrix}\right.\)
Ta có: \(P=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a}{2018-a}+\dfrac{b}{2018-b}+\dfrac{c}{2018-c}\)
\(P+3=\left(\dfrac{a}{2018-a}+1\right)+\left(\dfrac{b}{2018-b}+1\right)+\left(\dfrac{c}{2018-c}+1\right)=\dfrac{2018}{b+c}+\dfrac{2018}{c+a}+\dfrac{2018}{a+b}=2018\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+c}\right)=2018.\dfrac{2017}{2018}=2017\Rightarrow P=2014\)
Ta có : \(P=\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{b+a}\)
\(\Rightarrow3+P=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{a+c}+1\right)+\left(\dfrac{c}{a+b}+1\right)\)
\(\Rightarrow3+P=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a +b+c}{a+b}\)
\(\Rightarrow3+P=\left(a+b+c\right).\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)\)
Mà \(a+b+c=2018;\) \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{2017}{2018}\) \(\left(a,b\in R\right)\)
\(\Rightarrow3+P=2018.\dfrac{2017}{2018}\)
\(\Rightarrow3+P=2017\)
\(\Rightarrow P=2014\)
Vậy \(P=2014\)