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a. Vì \(\left|x-y-5\right|\ge0\forall x;y;2019\left|y-3\right|^{2020}\ge0\forall y\)
\(\Rightarrow\left|x-y-5\right|+2019\left|y-3\right|^{2020}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left|x-y-5\right|=0\\2019\left|y-3\right|^{2020}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y-5=0\\y-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y=5\\y=3\end{cases}}\)
b. \(2\left(x-5\right)^4\ge0\forall x;5\left|2y-7\right|^5\ge0\forall y\)
\(\Rightarrow2\left(x-5\right)^4+5\left|2y-7\right|^5\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}2\left(x-5\right)^4=0\\5\left|2y-7\right|^5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-5=0\\2y-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\y=\frac{7}{2}\end{cases}}\)
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\(M=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right).2.3.4...2018\)
\(\Rightarrow M=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right).2.3.4...673.674...2018\)
Vì \(\hept{\begin{cases}M⋮3\\M⋮673\end{cases}}\) mà \(\left(3,673\right)=1\) nên \(M⋮2019\left(đpcm\right)\)
\(M=\left[\left(1+\frac{1}{2018}\right)+\left(\frac{1}{2}+\frac{1}{2017}\right)+...+\left(\frac{1}{1008}+\frac{1}{1011}\right)+\left(\frac{1}{1009}+\frac{1}{1010}\right)\right].\)\(2.3...1008.1009.1010.1011...2017.2018\)
\(=\left(\frac{2019}{2018}+\frac{2019}{2.2017}+...+\frac{2019}{1008.1011}+\frac{2019}{1009.1010}\right).2.3...1008.1009.1010.1011...2017.2018\)
\(=2019\left(\frac{1}{2018}+\frac{1}{2.2017}+...+\frac{1}{1008.1011}+\frac{1}{1009.1010}\right).2...1008.1009.1010.1011...2017.2018\)
\(=2019.\left(2...2017+3...2016.2018+...+2.3...1007.1009.1011...2018+2.3....1008.1011...2018\right)\)
Chia hết cho 2019
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a) \(\left(4x-13\right)^4+4^3=145\)
\(\Rightarrow\left(4x-13\right)^4+64=145\)
\(\Rightarrow\left(4x-13\right)^4=81\)
\(\Rightarrow4x-13=\pm3\)
+) \(4x-13=3\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
+) \(4x-13=-3\)
\(\Rightarrow4x=10\)
\(\Rightarrow x=\frac{5}{2}\)
Vậy \(x=4\) hoặc \(x=\frac{5}{2}\)
b) \(3^{x+2}-3^x=72\)
\(\Rightarrow3^x.3^2-3^x=72\)
\(\Rightarrow3^x.\left(3^2-1\right)=72\)
\(\Rightarrow3^x.8=72\)
\(\Rightarrow3^x=9\)
\(\Rightarrow3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
c) \(2^{x+2}-2^{x-1}=224\)
\(\Rightarrow2^{x-1+3}-2^{x-1}=224\)
\(\Rightarrow2^{x-1}.2^3-2^{x-1}=224\)
\(\Rightarrow2^{x-1}.\left(2^3-1\right)=224\)
\(\Rightarrow2^{x-1}.7=224\)
\(\Rightarrow2^{x-1}=32\)
\(\Rightarrow2^{x-1}=2^5\)
\(\Rightarrow x-1=5\)
\(\Rightarrow x=6\)
Vậy x = 6