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Ta có: \(C=\frac{\left|x-2019\right|+2020}{\left|x-2019\right|+2021}=\frac{\left|x-2019\right|+2021-1}{\left|x-2019\right|+2021}=1-\frac{1}{\left|x-2019\right|+2021}\)
=> C đạt giá trị nhỏ nhất khi \(\frac{1}{\left|x-2019\right|+2021}\) lớn nhất
=> |x - 2019| + 2021 nhỏ nhất
Ta có: \(\left|x-2019\right|\ge0\)
\(\Rightarrow\left|x-2019\right|+2021\ge2021\)
Dấu "=" xảy ra khi x - 2019 = 0
=> x = 2019
\(\Rightarrow C=\frac{\left|2019-2019\right|+2020}{\left|2019-2019\right|+2021}=\frac{2020}{2021}\)
Vậy \(MinC=\frac{2020}{2021}\Leftrightarrow x=2019\).
Ta có:\(\frac{3-x}{2021}+\frac{2020-x}{2019}+\frac{4033-x}{2017}+\frac{6042-x}{2015}=10\)
\(\Leftrightarrow\frac{3-x}{2021}-1+\frac{2020-x}{2019}-2+\frac{4033-x}{2017}-3+\frac{6042-x}{2015}-4=0\)
\(\Leftrightarrow\frac{3-x-2021}{2021}+\frac{2020-x-4038}{2019}+\frac{4033-x-6051}{2017}+\frac{6042-x-8060}{2015}=0\)
\(\Leftrightarrow\frac{-2018-x}{2021}+\frac{-2018-x}{2019}+\frac{-2018-x}{2017}+\frac{-2018-x}{2015}=0\)
\(\Leftrightarrow-\left(2018+x\right)\left(\frac{1}{2021}+\frac{1}{2019}+\frac{1}{2017}+\frac{1}{2015}\right)=0\)
\(\Leftrightarrow2018+x=0.Do\frac{1}{2021}+\frac{1}{2019}+\frac{1}{2017}+\frac{1}{2015}>0\)
\(\Leftrightarrow x=-2018\)
V...
A(1/2^2022)=1/2^2022+1/2^4044+...+1/2^(2022^2021)
=>2^2022*A=1+1/2^2022+...+1/2^(2022^2020)
=>A*(2^2022-1)=1-1/2^(2022^2021)
=>\(A=\dfrac{2^{2022^{2021}}-1}{2^{2022}-1}\)
Ta có:
\(\frac{xy}{x+y}=\frac{yz}{y+z}=\frac{zx}{z+x}\rightarrow\frac{x+y}{xy}=\frac{y+z}{yz}=\frac{z+x}{zx}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{y}+\frac{1}{z}=\frac{1}{z}+\frac{1}{x}\Rightarrow\frac{1}{x}=\frac{1}{y}=\frac{1}{z}\Rightarrow x=y=z\)
Thay tất cả giá trị x,y,z vào M ta được:
\(M=\frac{2020x^3+2020y^3+2020z^3}{x^3+y^3+z^3}+\frac{2021x^5+2021y^5}{x^5+y^5}\)
\(\Rightarrow M=\frac{2020\left(x^3+y^3+z^3\right)}{x^3+y^3+z^3}+\frac{2021\left(x^5+y^5\right)}{x^5+y^5}\)
\(\Rightarrow M=2020+2021=4041\)
Ta có: \(B=\frac{3^{2021}-3^{2019}}{3^{2021}+3^{2020}}\)
\(\Leftrightarrow B=\frac{3^{2019}.\left(3^2-1\right)}{3^{2020}.\left(3+1\right)}\)
\(\Leftrightarrow B=\frac{8}{3.4}\)
\(\Leftrightarrow B=\frac{2}{3}\)
Vậy \(B=\frac{2}{3}\)
Ta có : \(B=\frac{3^{2021}-3^{2019}}{3^{2021}+3^{2020}}\)
\(=\frac{3^{2019}\left(3^2-1\right)}{3^{2020}\left(3+1\right)}\)
\(=\frac{3^{2019}.8}{3^{2020}.4}\)
\(=\frac{2}{3}\)