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So sánh :
A = \(\frac{2018^{2019}+1}{2018^{2020}+1}\)và B = \(\frac{2018^{2018}+1}{2018^{2019}+1}\)
Ta có :
+ , A = \(\frac{2018^{2019}+1}{2018^{2020}+1}\)
2018A = \(\frac{2018\left(2018^{2019}+1\right)}{2018^{2020}+1}\)
2018A = \(\frac{2018^{2020}+2018}{2018^{2020}+1}\)
2018A = \(1\frac{2018}{2018^{2020}+1}\)
+ , B = \(\frac{2018^{2018}+1}{2018^{2019}+1}\)
2018B = \(\frac{2018\left(2018^{2018}+1\right)}{2018^{2019}+1}\)
2018B = \(\frac{2018^{2019}+2018}{2018^{2019}+1}\)
2018B = \(1\frac{2018}{2018^{2019}+1}\)
Ta thấy :
20182019 + 1 < 20182020 + 1
= > \(\frac{2018}{2018^{2019}+1}\)> \(\frac{2018}{2018^{2020}+1}\)
= > B > A
Vậy B > A
a) \(A=2+2^2+2^3+...+2^{2019}\)
\(\Rightarrow2A=2^2+2^3+...+2^{2020}\)
\(\Rightarrow2A-A=\left(2^2+...+2^{2020}\right)-\left(2+...+2^{2019}\right)\)
\(\Rightarrow A=2^{2020}-2\)
Ta có: \(A+2=2^{x+10}\)
\(\Leftrightarrow2^{2020}-2+2=2^{x+10}\)
\(\Leftrightarrow2^{2020}=2^{x+10}\)
\(\Leftrightarrow2020=x+10\)
\(\Leftrightarrow x=2010\)
b) Ta có: \(A+2=2^{2020}=\left(2^{1010}\right)^2\)là số chính phương
XÉT:\(A=2+2^2+2^3+...+2^{2019}\)
\(\Leftrightarrow2A=2^2+2^3+...+2^{2019}+2^{2020}\)
\(\Leftrightarrow2A-A=2^{2020}-2\)
\(\Leftrightarrow A=2^{2020}-2\)
\(\Rightarrow A+2=2^{2020}-2+2=2^{2020}\)LÀ SỐ CHÍNH PHƯƠNG
MÀ\(a+2=2^{x+10}\)
\(\Leftrightarrow2^{x+10}=2^{2020}\)
\(\Leftrightarrow x+10=2020\Leftrightarrow x=2010\)
a) (x-3)100-1=0
\(\Leftrightarrow\left(x-3\right)^{100}=0+1=1\)
\(\Leftrightarrow\left(x-3\right)^{100}=1^{100}=\left(-1\right)^{100}\)
\(\Rightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1+3=4\\x=\left(-1\right)+3=2\end{matrix}\right.\)
Vậy: x\(\in\left\{4;2\right\}\)
b) (9-5x)2019+1=0
\(\Leftrightarrow\left(9-5x\right)^{2019}=0-1=-1=\left(-1\right)^{2019}\)
\(\Rightarrow9-5x=-1\)
\(\Leftrightarrow5x=9-\left(-1\right)=10\)
\(\Leftrightarrow x=10:5=2\)
Vậy: x=2
Ta có : 2019^10+2019^9=2019^9.(2019+1)=2019^9.2020
Mà 2020^10>2019^9.2020
=>2020^10>2019^10+2019^9
\(A=\frac{1}{2^3}+\frac{1}{3^3}+...+\frac{1}{2019^3}\)
\(\Rightarrow A< \frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{2018\cdot2019\cdot2020}\)
\(\Rightarrow A< \frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\right)\)
\(\Rightarrow A< \frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{2019\cdot2020}\right)\)
\(\Rightarrow A< \frac{1}{4}-\left(\frac{1}{2}\cdot\frac{1}{2019\cdot2020}\right)\)
\(\Rightarrow A< \frac{1}{4}\) ( ĐPCM )
\(A=1+2019+2019^2+...+2019^{50}\\ 2019A=2019+2019^2+2019^3+...+2019^{51}\\ 2019A-A=2019^{51}-1\\ \Rightarrow A=\frac{2019^{51}-1}{2018}\)
Nhanh thật =))