Cho 3 số a,b,c thỏa mãn : abc = 2018
Tính M = \(\frac{2018a}{ab+2018a+2018}+\frac{b}{bc+b+2018}+\frac{c}{ac+c+1}\)
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10 tháng 5 2018
M = \(\dfrac{2018a}{ab+2018a+2018}+\dfrac{b}{bc+b+2018}+\dfrac{c}{ac+c+1}\)
M = \(\dfrac{a^2bc}{ab+a^2bc+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}\)
M = \(\dfrac{a^2bc}{ab\left(ac+c+1\right)}+\dfrac{b}{b\left(ac+c+1\right)}+\dfrac{c}{ac+c+1}\)
M= \(\dfrac{ac}{ac+c+1}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}\)
M = \(\dfrac{ac+c+1}{ac+c+1}\)
M = 1
22 tháng 1 2019
\(\frac{2018}{ab+2018a+2018}+\frac{b}{bc+a+2018}+\frac{c}{ac+c+1}\)
\(a.b.c=2018\Rightarrow a,b,c\ne0\)
Ta có \(\frac{2018}{ab+2018a+2018}\Rightarrow\frac{2018}{b+2018+bc}\)
\(\frac{c}{ac+c+1}=\frac{bc}{abc+bc+b}=\frac{bc}{2018+bc+b}\)
\(\Rightarrow S=\frac{2018}{b+2018+bc}+\frac{b}{bc+b+2018}+\frac{bc}{2018+bc+b}=\frac{2018+b+bc}{b+2018+bc}=1\)
để nghĩ tiếp
22 tháng 1 2019
làm tiếp
\(\frac{2013x+1}{2014x-2014}=\frac{2013\left(x-1\right)+2014}{2014\left(x-1\right)}=\frac{2013}{2014}+\frac{1}{x-1}\)
\(B_{max}\Leftrightarrow\frac{1}{x-1}max\)
+) Nếu x >1 thì x-1 >0 \(\Rightarrow\frac{1}{x-1}>0\)
+) Nếu x<1 thì x-1 <0 \(\Rightarrow\frac{1}{x-1}< 0\)
Xét x > 1 ta có
\(\frac{1}{x-1}max\Rightarrow x-1\)là số nguyên dương nhỏ nhất
\(\Rightarrow x-1=1\Rightarrow x=2\)
Vậy \(Bmax=1\frac{2018}{2019}\Leftrightarrow x=2\)
E
11 tháng 12 2018
Số 2018 kia như đang gợi ý cho bài này vậy :)
\(M=\dfrac{a^2bc}{ab+a^2bc+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}=\dfrac{a^2bc}{ab\left(ac+c+1\right)}+\dfrac{b}{b\left(ac+c+1\right)}+\dfrac{c}{ac+c+1}=\dfrac{ac}{ac+c+1}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}=\dfrac{ac+c+1}{ac+c+1}=1\)
15 tháng 1 2019
Có: \(\frac{2018a+3}{1+b^2}=2018a+3-\frac{b^2\left(2018a+3\right)}{1+b^2}\) (Làm tắt ráng hiểu ^^)
\(\ge2018a+3-\frac{b^2\left(2018a+3\right)}{2b}\left(Cauchy\right)\)
\(=2018a+3-\frac{b\left(2018a+3\right)}{2}\)
\(=2018a+3-\frac{2018ab+3b}{2}\)
Tương tự \(\frac{2018b+3}{1+c^2}\ge2018b+3-\frac{2018bc+3b}{2}\)
\(\frac{2018c+3}{1+a^2}\ge2018c+3-\frac{2018ac+3a}{2}\)
CỘng vế với vế của các bđt trên lại ta được
\(A\ge2018\left(a+b+c\right)+9-\frac{2018\left(ab+bc+ca\right)+3\left(a+b+c\right)}{2}\)
\(=2018\left(a+b+c\right)+9-\frac{6054+3\left(a+b+c\right)}{2}\)
\(=2018\left(a+b+c\right)-\frac{3\left(a+b+c\right)}{2}-3018\)
\(=\frac{4033\left(a+b+c\right)}{2}-3018\)
Ta có bđt phụ : \(a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}\)(1)
Thật vậy \(\left(1\right)\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc\ge3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(Luôn đúng)
Nên (1) được chứng minh
ÁP dụng (1) ta được \(A\ge\frac{4033\left(a+b+c\right)}{2}-3018\ge\frac{4033}{2}\sqrt{3\left(ab+bc+ca\right)}-3018\)
\(=\frac{4033}{2}\sqrt{3.3}-3018\)
\(=\frac{6063}{2}\)
Dấu "='' xảy ra \(\Leftrightarrow\hept{\begin{cases}a=b=c\\ab+bc+ca=3\end{cases}\Leftrightarrow}a=b=c=1\)
Vậy \(A_{min}=\frac{6063}{2}\Leftrightarrow a=b=c=1\)
HP
17 tháng 8 2021
\(P=\dfrac{a}{a+\sqrt{2018a+bc}}+\dfrac{b}{b+\sqrt{2018b+ca}}+\dfrac{c}{c+\sqrt{2018c+ab}}\)
\(=\dfrac{a}{a+\sqrt{a.\left(a+b+c\right)+bc}}+\dfrac{b}{b+\sqrt{b.\left(a+b+c\right)+ca}}+\dfrac{c}{c+\sqrt{c.\left(a+b+c\right)+ab}}\)
\(=\dfrac{a}{a+\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{b+\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{c+\sqrt{c^2+ab+bc+ca}}\)
\(=\dfrac{a\left(\sqrt{a^2+ab+bc+ca}-a\right)}{ab+bc+ca}+\dfrac{b\left(\sqrt{b^2+ab+bc+ca}-b\right)}{ab+bc+ca}+\dfrac{c\left(\sqrt{c^2+ab+bc+ca}-c\right)}{ab+bc+ca}\)
\(=\dfrac{a\left(\sqrt{\left(a+b\right)\left(a+c\right)}-a\right)}{ab+bc+ca}+\dfrac{b\left(\sqrt{\left(b+c\right)\left(b+a\right)}-b\right)}{ab+bc+ca}+\dfrac{c\left(\sqrt{\left(c+a\right)\left(c+b\right)}-c\right)}{ab+bc+ca}\)
\(\le\dfrac{a\left(\dfrac{2a+b+c}{2}-a\right)}{ab+bc+ca}+\dfrac{b\left(\dfrac{2b+c+a}{2}-b\right)}{ab+bc+ca}+\dfrac{c\left(\dfrac{2c+b+a}{2}-c\right)}{ab+bc+ca}\)
\(=\dfrac{ab+ac}{2\left(ab+bc+ca\right)}+\dfrac{bc+ba}{2\left(ab+bc+ca\right)}+\dfrac{ca+cb}{2\left(ab+bc+ca\right)}\)
\(=\dfrac{2\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=1\)
\(maxP=1\Leftrightarrow a=b=c=\dfrac{2018}{3}\)
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 ...
\(M=\frac{2018a}{ab+2018a+2018}+\frac{b}{bc+b+2018}+\frac{c}{ac+c+1}\)
\(\Rightarrow M=\frac{2018a}{ab+2018a+2018}+\frac{ab}{a\left(bc+b+2018\right)}+\frac{abc}{ab\left(ac+c+1\right)}\)
\(\Rightarrow M=\frac{2018a}{ab+2018a+2018}+\frac{ab}{ab+2018a+2018}+\frac{1}{ab+2018a+2018}\)
\(\Rightarrow M=\frac{2018a+ab+1}{2018a+ab+1}=1\)
Do : \(abc=2018\)nên : \(a,b,c\ne0\)
Ta có : \(M=\frac{2018a}{ab+2018a+2018}+\frac{b}{bc+b+2018}+\frac{c}{ac+c+1}\)
\(=\frac{2018a}{ab+2018a+2018}+\frac{ab}{abc+ab+2018a}+\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{2018a}{ab+2018a+2018}+\frac{ab}{2018+ab+2018a}+\frac{2018}{2018+ab+2018a}\)
\(=\frac{2018a+ab+2018}{ab+2018a+2018}=1\)