cho biểu thức: M= \(2^{2018}+2^{2020}\)
chứng minh rằng: M\(⋮\)5120
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NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 2 2024
\(\sqrt{1+\dfrac{1}{x^2}+\dfrac{1}{\left(x+1\right)^2}}=\sqrt{\dfrac{x^2+\left(x+1\right)^2+x^2\left(x+1\right)^2}{x^2\left(x+1\right)^2}}=\sqrt{\dfrac{x^2\left(x+1\right)^2+2x^2+2x+1}{x^2\left(x+1\right)^2}}\)
\(=\sqrt{\dfrac{\left(x^2+x\right)^2+2\left(x^2+x\right)+1}{\left(x^2+x\right)^2}}=\sqrt{\dfrac{\left(x^2+x+1\right)^2}{\left(x^2+x\right)^2}}=\dfrac{x^2+x+1}{x^2+x}\)
\(=1+\dfrac{1}{x}-\dfrac{1}{x+1}\)
\(\Rightarrow f\left(1\right).f\left(2\right)...f\left(2020\right)=5^{1+1-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{2020}-\dfrac{1}{2021}}\)
\(=5^{2021-\dfrac{1}{2021}}\)
\(\Rightarrow\dfrac{m}{n}=2021-\dfrac{1}{2021}=\dfrac{2021^2-1}{2021}\)
\(\Rightarrow m-n^2=2021^2-1-2021^2=-1\)
2 tháng 1 2023
\(5x^2+5y^2+8xy-2x+2y+2=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
Vì \(\left(x+y\right)^2\ge0,\left(x-1\right)^2\ge0,\left(y+1\right)^2\ge0\)
\(\Rightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(\left(x+y\right)^{2018}+\left(x-2\right)^{2019}+\left(y+1\right)^{2020}=\left(1-1\right)^{2018}+\left(1-2\right)^{2019}+\left(-1+1\right)^{2020}=-1\)
20 tháng 11 2022
Bài 1:
Đặt 2018=a
\(B=\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\)
\(=1+a-\dfrac{a}{a+1}+\dfrac{a}{a+1}=1+a=2019\)
L
10 tháng 5 2018
Ta có: \(a^{2017}+b^{2017}\)= \(2a^{^{ }1018}.b^{1018}\)
⇔ (a2017 + b2017)2 = 4(ab)2018
Lại có: (a2017 + b2017)2 ≥ 4a2017.b2017
⇒ 4(ab)2016 ≥ 4a2017.b2017
⇒ ab2016 ≥ ab2017
⇒ ab ≤ 1
⇒ 1 - ab ≥ 0
⇒ 2018 - 2018ab ≥ 0
\(M=2^{2018}+2^{2020}=2^{2018}.\left(1+2^2\right)=2^{2018}.5=2^{2008}.\left(2^{10}.5\right)=2^{2008}.\left(1024.5\right)=2^{2008}.5120⋮5120\)
\(2^{2018}+2^{2020}\)
\(=2^{2018}\left(1+2^2\right)\)
\(=2^{2018}.5\)
\(=2^{2010}.5120⋮5120\)
\(\RightarrowĐPCM\)