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\(\frac{2\left|2018x-2019\right|+2019}{\left|2018x-2019\right|+1}\)
\(=\frac{\left(2\left(\left|2018x-2019\right|+1\right)\right)+2017}{\left|2018x-2019\right|+1}\)
\(=2+\frac{2017}{\left|2018x-2019\right|+1}\)có giá trị lớn nhất
\(\Rightarrow\frac{2017}{\left|2018x-2019\right|+1}\)có giá trị lớn nhất
\(\Rightarrow\left|2018x-2019\right|+1\)có giá trị nhỏ nhất
Mà \(\left|2018x-2019\right|\ge0\)
\(\Rightarrow\left|2018x-2019\right|+1\ge1\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left|2018x-2019\right|=0\)
\(\Leftrightarrow x=\frac{2019}{2018}\)
Vậy \(M_{MAX}=2019\)tại \(x=\frac{2019}{2018}\)
\(\frac{5^x+5^{x+1}+5^{x+2}}{31}=\frac{3^{2x}+3^{2x+1}+3^{2x+2}}{13}\)
\(\Rightarrow\frac{5^x\left(1+5+5^2\right)}{31}=\frac{3^{2x}\left(1+3+3^2\right)}{13}\)
\(\Rightarrow\frac{5^x\cdot31}{31}=\frac{3^{2x}\cdot13}{13}\)
\(\Rightarrow5^x=3^{2x}\)
Mà \(\left(5;3\right)=1\)
\(\Rightarrow x=2x=0\)
ta có \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{b+c}\)
=\(\frac{a}{b+c}+1+\frac{b}{a+c}+1+\frac{c}{b+c}+1-3\)
=\(\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\)
=\(\left(a+b+c\right)\left(\frac{1}{c+b}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)
rồi còn lại thay vào nha bn
\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2019\cdot\frac{1}{2019}\)
\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=1\)
\(\Leftrightarrow\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)+3=1\)
\(\Leftrightarrow S=-2\)
Có: \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{2019}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2019.\frac{1}{2019}\)
\(\Leftrightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{a+c}=1\)
\(\Leftrightarrow\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{a+c}=-2\)
Ta có : \(P=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(\Rightarrow P+3=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\)
\(\Rightarrow P+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)
\(\Rightarrow P+3=\left(a+b+c\right).\frac{1}{b+c}+\left(a+b+c\right).\frac{1}{c+a}+\left(a+b+c\right).\frac{1}{a+b}\)
\(\Rightarrow P+3=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)
\(\Rightarrow P+3=2019.10\)
\(\Rightarrow P+3=20190\)
\(\Rightarrow P=20190-3\)
\(\Rightarrow P=20187\)
Vậy P = 20187
Ta có S + 4 = \(\left(\frac{a}{b+c+d}+1\right)+\left(\frac{b}{c+d+a}+1\right)+\left(\frac{c}{a+b+d}+1\right)+\left(\frac{d}{a+b+c}+1\right)\)
\(=\frac{a+b+c+d}{b+c+d}+\frac{a+b+c+d}{a+c+d}+\frac{a+b+c+d}{a+b+d}+\frac{a+b+c+d}{b+c+d}\)
\(=\left(a+b+c+d\right)\left(\frac{1}{b+c+d}+\frac{1}{a+c+d}+\frac{1}{a+b+d}+\frac{1}{a+b+c}\right)\)
\(=4000.\frac{1}{40}=100\)(a + b + c + d = 4000 ; \(\frac{1}{b+c+d}+\frac{1}{a+c+d}+\frac{1}{a+b+d}+\frac{1}{a+b+c}=\frac{1}{40}\))
=> S = 100 - 4 = 96
theo bài ra ta có
\(\frac{a^{2015}}{b^{2017}+c^{2019}}=\frac{b^{2017}}{a^{2015}+c^{2019}}=\frac{c^{2019}}{a^{2015}+b^{2017}}\)
=>\(\frac{a^{2015}}{b^{2017}+c^{2019}}+1=\frac{b^{2017}}{a^{2015}+c^{2019}}+1=\frac{c^{2019}}{a^{2015}+b^{2017}}+1\)
=> \(\frac{a^{2015}+b^{2017}+c^{2019}}{b^{2017}+c^{2019}}=\frac{a^{2015}+b^{2017}+c^{2019}}{a^{2015}+c^{2019}}=\frac{a^{2015}+b^{2017}+c^{2019}}{a^{2015}+b^{2017}}\)
- nếu a2015+ b2017 +c2019 = 0
=> b2017+ c2019 = -(a2015) (1)
=> a2015+ c2019= -(b2017) (2)
=> a2015+ b2017= -(c2019) (3)
thay 1, 2, 3 vào S ta có:
S = \(\frac{b^{2017}+c^{2019}}{a^{2015}}+\frac{a^{2015}+c^{2019}}{b^{2017}}+\frac{a^{2015}+b^{2017}}{c^{2019}}\)
=> S =\(\frac{-\left(a^{2015}\right)}{a^{2015}}+\frac{-\left(b^{2017}\right)}{b^{2017}}+\frac{-\left(c^{2019}\right)}{c^{2019}}\)
S = -1 + -1 + -1
S = -3
vậy S ko phụ thuộc vào giá trị a,b,c
- nếu a2015+b2017+c2019 khác 0
=> b2017+c2019 = a2015+c2019=a2015+b2017
=> b2017 = a2015 = c2019
=>S=\(\frac{b^{2017}+c^{2019}}{a^{2015}}+\frac{a^{2015}+c^{2019}}{b^{2017}}+\frac{a^{2015}+b^{2017}}{c^{2019}}=\frac{2a^{2015}}{a^{2015}}+\frac{2b^{2017}}{b^{2017}}+\frac{2c^{2019}}{c^{2019}}=2+2+2=6\)
VẬY S ko phụ thuộc vào các giá trị của a,b,c
từ 2 trường hợp trên => giá trị của biểu thức S ko phụ thuộc vào giá trị của a,b,c (đpcm)
Ta có:\(\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\left(a+b+c\right)=\frac{1}{3}.2028\)
=>\(\left(\frac{a+b}{a+b}+\frac{c}{a+b}\right)+\left(\frac{b+c}{b+c}+\frac{a}{b+c}\right)+\left(\frac{c+a}{c+a}+\frac{b}{c+a}\right)=676\)
=>\(\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}+3=676\)
=>\(Q=673\)
Vậy Q=673
dự đoán của chúa Pain
a=b=c=\(\frac{2028}{3}\)
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{2\left(a+b+c\right)}\left(cosi\right).\)
\(Q\ge\frac{\left(a+b+c\right)}{2\left(a+b+c\right)}+\frac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}{2\left(a+b+c\right)}\)
\(Q\ge\frac{1}{2}+\frac{\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}{\left(a+b+c\right)}\)
có
\(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge3\sqrt[3]{\sqrt{a^2b^2c^2}}=3\sqrt[3]{abc}\)
có
\(a+b+c\ge3\sqrt[3]{abc}\)
thay vào ta được
\(Q\ge\frac{1}{2}+\frac{3\sqrt[3]{abc}}{3\sqrt[3]{abc}}=\frac{1}{2}+1=\frac{3}{2}\)
dấu = xảy ra khi \(a=b=c=\frac{2028}{3}=676\)
thử thay vào ta được
\(Q=\frac{676}{2\left(676\right)}+\frac{676}{2\left(676\right)}+\frac{676}{2\left(676\right)}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}\) ( đúng )
