Cho ba số a,b,c thỏa mãn abc=2018. Tính giá trị biểu thức
M= \(\dfrac{2018a}{ab+2018a+2018}+\dfrac{b}{bc+b+2018}+\dfrac{c}{ac+c+1}\)
cho abc=2018
tính giá trị của biểu thức M=\(\dfrac{2018a}{ab+2018a+2018}+\dfrac{b}{bc+b+2018}+\dfrac{c}{ac+c+1}\)
Số 2018 kia như đang gợi ý cho bài này vậy :)
\(M=\dfrac{a^2bc}{ab+a^2bc+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}=\dfrac{a^2bc}{ab\left(ac+c+1\right)}+\dfrac{b}{b\left(ac+c+1\right)}+\dfrac{c}{ac+c+1}=\dfrac{ac}{ac+c+1}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}=\dfrac{ac+c+1}{ac+c+1}=1\)
cho abc=2018
tính giá trị của biểu thức M=\(\dfrac{2018a}{ab+2018a+2018}+\dfrac{b}{bc+b+2018}+\dfrac{c}{ac+c+1}\)
Cho 3 số a,b,c thỏa mãn : abc = 2018
Tính M = \(\frac{2018a}{ab+2018a+2018}+\frac{b}{bc+b+2018}+\frac{c}{ac+c+1}\)
\(M=\frac{2018a}{ab+2018a+2018}+\frac{b}{bc+b+2018}+\frac{c}{ac+c+1}\)
\(\Rightarrow M=\frac{2018a}{ab+2018a+2018}+\frac{ab}{a\left(bc+b+2018\right)}+\frac{abc}{ab\left(ac+c+1\right)}\)
\(\Rightarrow M=\frac{2018a}{ab+2018a+2018}+\frac{ab}{ab+2018a+2018}+\frac{1}{ab+2018a+2018}\)
\(\Rightarrow M=\frac{2018a+ab+1}{2018a+ab+1}=1\)
Do : \(abc=2018\)nên : \(a,b,c\ne0\)
Ta có : \(M=\frac{2018a}{ab+2018a+2018}+\frac{b}{bc+b+2018}+\frac{c}{ac+c+1}\)
\(=\frac{2018a}{ab+2018a+2018}+\frac{ab}{abc+ab+2018a}+\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{2018a}{ab+2018a+2018}+\frac{ab}{2018+ab+2018a}+\frac{2018}{2018+ab+2018a}\)
\(=\frac{2018a+ab+2018}{ab+2018a+2018}=1\)
cho a,b,c dương thỏa mãn điều kiện a+b+c=2018.tìm GTLN
\(P=\dfrac{a}{a+\sqrt{2018a+bc}}+\dfrac{b}{b+\sqrt{2018b+ca}}+\dfrac{c}{c+\sqrt{2018c+ab}}\)
cho a,b,c dương thỏa mãn điều kiện a+b+c=2018.tìm GTLN
\(P=\dfrac{a}{a+\sqrt{2018a+bc}}+\dfrac{b}{b+\sqrt{2018b+ca}}+\dfrac{c}{c+\sqrt{2018c+ab}}\)
\(P=\dfrac{a}{a+\sqrt{2018a+bc}}+\dfrac{b}{b+\sqrt{2018b+ca}}+\dfrac{c}{c+\sqrt{2018c+ab}}\)
\(=\dfrac{a}{a+\sqrt{a.\left(a+b+c\right)+bc}}+\dfrac{b}{b+\sqrt{b.\left(a+b+c\right)+ca}}+\dfrac{c}{c+\sqrt{c.\left(a+b+c\right)+ab}}\)
\(=\dfrac{a}{a+\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{b+\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{c+\sqrt{c^2+ab+bc+ca}}\)
\(=\dfrac{a\left(\sqrt{a^2+ab+bc+ca}-a\right)}{ab+bc+ca}+\dfrac{b\left(\sqrt{b^2+ab+bc+ca}-b\right)}{ab+bc+ca}+\dfrac{c\left(\sqrt{c^2+ab+bc+ca}-c\right)}{ab+bc+ca}\)
\(=\dfrac{a\left(\sqrt{\left(a+b\right)\left(a+c\right)}-a\right)}{ab+bc+ca}+\dfrac{b\left(\sqrt{\left(b+c\right)\left(b+a\right)}-b\right)}{ab+bc+ca}+\dfrac{c\left(\sqrt{\left(c+a\right)\left(c+b\right)}-c\right)}{ab+bc+ca}\)
\(\le\dfrac{a\left(\dfrac{2a+b+c}{2}-a\right)}{ab+bc+ca}+\dfrac{b\left(\dfrac{2b+c+a}{2}-b\right)}{ab+bc+ca}+\dfrac{c\left(\dfrac{2c+b+a}{2}-c\right)}{ab+bc+ca}\)
\(=\dfrac{ab+ac}{2\left(ab+bc+ca\right)}+\dfrac{bc+ba}{2\left(ab+bc+ca\right)}+\dfrac{ca+cb}{2\left(ab+bc+ca\right)}\)
\(=\dfrac{2\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=1\)
\(maxP=1\Leftrightarrow a=b=c=\dfrac{2018}{3}\)
Cho a,b,c thỏa mãn a.b.c=2018
tính S= 2018/ab+2018a+2018 +b/bc+b+2018 +c/ac+c+1
Tìm x thuộc Z để B=2013x+1/2014x-2014 đạt GTLN
\(\frac{2018}{ab+2018a+2018}+\frac{b}{bc+a+2018}+\frac{c}{ac+c+1}\)
\(a.b.c=2018\Rightarrow a,b,c\ne0\)
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
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 \(Bmax=1\frac{2018}{2019}\Leftrightarrow x=2\)
Cho các số thực dương a,b,c thỏa mãn a+b+c=2018. Tìm GTNN của
P = \(\dfrac{a}{a+\sqrt{2018a+bc}}+\dfrac{b}{b+\sqrt{2108b+ca}}+\dfrac{c}{c+\sqrt{2018c+ab}}\)
Ap dung BDT Cauchy-Schwarz ta co:
\(\dfrac{a}{a+\sqrt{2018a+bc}}=\dfrac{a}{a+\sqrt{a\left(a+b+c\right)+bc}}\)
\(=\dfrac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\ge\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)
Tuong tu cho 2 BDT con lai roi cong theo ve:
\(P\ge\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\)
Cho a, b, c thỏa mãn (a + b + c)(ab + bc + ac) = 2018 và abc = 2018. Tính giá trị của biểu thức P = (b^2.c + 2018)(a^2.b + 2018)(c^2.a + 2018)
cho ba số thực dương a b c thỏa mãn ab+bc+ac≤1. tìm giá trị nhỏ nhất của biểu thức P biết:
P= \(\dfrac{1}{\sqrt{a^2+b^2-abc}}+\dfrac{1}{\sqrt{a^2+c^2-abc}}+\dfrac{1}{\sqrt{c^2+b^2-abc}}\)