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C với D mình làm sau vì nó phức tạp hơn ... E với F trước nhé
E = | 3x + 1 | + 2| x - y | + 1
\(\hept{\begin{cases}\left|3x+1\right|\ge0\\2\left|x-y\right|\ge0\end{cases}\forall}x,y\Rightarrow\left|3x+1\right|+2\left|x-y\right|+1\ge1\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}3x+1=0\\x-y=0\end{cases}}\Leftrightarrow x=y=-\frac{1}{3}\)
=> MinE = 1 <=> x = y = -1/3
F = 5| x - 1 | + 1/2| 2x + y | + 2020
\(\hept{\begin{cases}5\left|x-1\right|\ge0\\\frac{1}{2}\left|2x+y\right|\ge0\end{cases}\forall}x,y\Rightarrow5\left|x-1\right|+\frac{1}{2}\left|2x+y\right|+2020\ge0\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-1=0\\2x+y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
=> MinF = 2020 <=> x = 1 ; y = -2
C = 2| x - 1 | + | 2x + 3 | - 2020
= | 2x - 2 | + | 2x + 3 | - 2020
= | 2x - 2 | + | -( 2x + 3 ) | - 2020
= | 2x - 2 | + | -2x - 3 | - 2020
Áp dụng bất đẳng thức | a | + | b | ≥ | a + b | ta có :
C = | 2x - 2 | + | -2x - 3 | - 2020 ≥ | 2x - 2 - 2x - 3 | - 2020 = | -5 | - 2020 = 5 - 2020 = -2015
Dấu "=" xảy ra khi ab ≥ 0
=> ( 2x - 2 )( -2x - 3 ) ≥ 0
Xét hai trường hợp :
1. \(\hept{\begin{cases}2x-2\ge0\\-2x-3\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\ge2\\-2x\ge3\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\le-\frac{3}{2}\end{cases}}\)( loại )
2. \(\hept{\begin{cases}2x-2\le0\\-2x-3\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\le2\\-2x\le3\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge-\frac{3}{2}\end{cases}}\Leftrightarrow-\frac{3}{2}\le x\le1\)
=> MinC = -2015 <=> \(-\frac{3}{2}\le x\le1\)
D = | 3 - 2x | + 2| 1 - x | + 1/2
= | 3 - 2x | + | 2 - 2x | + 1/2
= | -( 3 - 2x ) | + | 2 - 2x | + 1/2
= | 2x - 3 | + | 2 - 2x | + 1/2
Áp dụng bất đẳng thức | a | + | b | ≥ | a + b | ta có :
D = | 2x - 3 | + | 2 - 2x | + 1/2 ≥ | 2x - 3 + 2 - 2x | + 1/2 = | -1 | + 1/2 = 1 + 1/2 = 3/2
Dấu "=" xảy ra khi ab ≥ 0
=> ( 2x - 3 )( 2 - 2x ) ≥ 0
Xét hai trường hợp :
1. \(\hept{\begin{cases}2x-3\ge0\\2-2x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\ge3\\-2x\ge-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\x\le1\end{cases}}\)( loại )
2. \(\hept{\begin{cases}2x-3\le0\\2-2x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\le3\\-2x\le-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{2}\\x\ge1\end{cases}}\Leftrightarrow1\le x\le\frac{3}{2}\)
=> MinD = 3/2 <=> \(1\le x\le\frac{3}{2}\)
Đặt \(A=1-\frac{1}{2^2}-\frac{1}{3^2}-.........-\frac{1}{2020^2}\)
Ta có: \(2^2=2.2< 2.3\)\(\Rightarrow\frac{1}{2.2}>\frac{1}{2.3}\)\(\Rightarrow\frac{1}{2^2}>\frac{1}{2.3}\)
Tương tự, ta có: \(\frac{1}{3^2}>\frac{1}{3.4}\), ........... , \(\frac{1}{2020^2}>\frac{1}{2020.2021}\)
\(\Rightarrow A>1-\frac{1}{2.3}-\frac{1}{3.4}-...........-\frac{1}{2020.2021}\)
\(=1-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{3}-\frac{1}{4}\right)-.......-\left(\frac{1}{2020}-\frac{1}{2021}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-..........-\frac{1}{2020}+\frac{1}{2021}\)
\(=1-\frac{1}{2}+\frac{1}{2021}\)\(=\frac{1}{2}+\frac{1}{2021}=\frac{2023}{4042}>\frac{1}{2020}\)
\(\Rightarrow A>\frac{1}{2020}\)
Đặt \(K=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2020}\)
\(=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{2020.2021}{2}}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2020.2021}\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2020}-\frac{1}{2021}\right)\)
\(=2\left(1-\frac{1}{2021}\right)=2.\frac{2020}{2021}=\frac{4040}{2021}\)
\(\Rightarrow D=\frac{2020}{\frac{4040}{2021}}=\frac{2021}{2}\)
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{2020}\left(1+2+3+...+2020\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{2020}.\frac{2020.2021}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{2021}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{2021}{2}\)
\(=\frac{\left[\left(2021-2\right)+1\right]\left(2021+2\right)}{2}:2\)
\(=1021615\)
Lời giải:
Đặt: \(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{2019^2}=a\).
Biểu thức $A$ lúc đó được biểu diễn như sau:
\(A=a(a-1+\frac{1}{2020^2})-(a+\frac{1}{2020^2})(a-1)\)
\(=a(a-1)+\frac{a}{2020^2}-[a(a-1)+\frac{a-1}{2020^2}]\)
\(=\frac{1}{2020^2}\)
Gọi biểu thức là A
3A= \(1+\frac{1}{3}+...+\frac{1}{3^{2019}}\)
⇒ 3A-A=2A=\(1+\frac{1}{3}+...+\frac{1}{3^{2019}}\)-\(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2020}}\)
⇒ 2A=1-\(\frac{1}{3^{2020}}\)
⇒ A= \(\frac{1}{2}-\frac{1}{3^{2020}.2}\)
⇒ A< \(\frac{1}{2}\)