Cho a,b,c là số thực thỏa mãn: (a+b+c)(ab+bc+ca)=2018 và abc=2018
Tính P= (b2c+2018)(c2a+2018)(a2b+2018)
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NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 1 2019
\(\left(a+b+c\right)\left(ab+ac+bc\right)=\left(a+b+c\right)\left(ab+ac+bc+c^2-c^2\right)\)
\(=\left(a+b+c\right)\left(\left(a+c\right)\left(b+c\right)-c^2\right)\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)-c^2\left(a+b\right)+c\left(a+c\right)\left(b+c\right)-c^3\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)-c^2a-c^2b+abc+c^2a+c^2b+c^3-c^3\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)+abc=\left(a+b\right)\left(a+c\right)\left(b+c\right)+2018\)
\(\Rightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)+2018=2018\)
\(\Rightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
Ta có:
\(A=\left(b^2c+2018\right)\left(c^2a+2018\right)\left(a^2b+2018\right)\)
\(A=\left(b^2c+abc\right)\left(c^2a+abc\right)\left(a^2b+abc\right)\)
\(A=bc\left(a+b\right)ac\left(b+c\right)ab\left(a+c\right)\)
\(A=\left(abc\right)^2\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(A=2018^2.0=0\)
21 tháng 1 2019
\(M=\frac{2018a}{ab+2018a+2018}+\frac{b}{bc+b+2018}+\frac{c}{ac+c+1}\)
\(\Rightarrow M=\frac{2018a}{ab+2018a+2018}+\frac{ab}{a\left(bc+b+2018\right)}+\frac{abc}{ab\left(ac+c+1\right)}\)
\(\Rightarrow M=\frac{2018a}{ab+2018a+2018}+\frac{ab}{ab+2018a+2018}+\frac{1}{ab+2018a+2018}\)
\(\Rightarrow M=\frac{2018a+ab+1}{2018a+ab+1}=1\)
X
21 tháng 1 2019
Do : \(abc=2018\)nên : \(a,b,c\ne0\)
Ta có : \(M=\frac{2018a}{ab+2018a+2018}+\frac{b}{bc+b+2018}+\frac{c}{ac+c+1}\)
\(=\frac{2018a}{ab+2018a+2018}+\frac{ab}{abc+ab+2018a}+\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{2018a}{ab+2018a+2018}+\frac{ab}{2018+ab+2018a}+\frac{2018}{2018+ab+2018a}\)
\(=\frac{2018a+ab+2018}{ab+2018a+2018}=1\)
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 \(\left(a;b;c\right)=\left(2.1;2.2017;2.2018\right)=...\)
TT
19 tháng 2 2020
Do \(abc=2018,bc+b+1\ne0\) nên thay vào biểu thức A ta có :
\(A=\frac{2018}{abc+bc+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+2018}\)
\(=\frac{abc}{a\left(bc+b+1\right)}+\frac{b}{bc+b+1}+\frac{a}{ab+a+abc}\)
\(=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{a}{a\left(bc+b+1\right)}\)
\(=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}\)
\(=\frac{bc+b+1}{bc+b+1}=1\)
Vậy : \(A=1\) với a,b,c thỏa mãn đề.
19 tháng 2 2020
\(A=\frac{2018}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+2018}\)
\(=\frac{abc}{abc+ab+a}+\frac{ab}{abc+ab+a}+\frac{a}{ab+a+abc}\)
\(=1\)
Vậy ...