Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:A= \(1+2+2^2+2^3+...+2^{2010}\)
=> 2A= 2(\(1+2+2^2+2^3+...+2^{2010}\))
=> 2A= 2 +\(2^2+2^3+2^4+...+2^{2011}\)
=> 2A-A= A =(2+ \(2^2+2^3+2^4+...+2^{2011}\)) -( \(1+2+2^2+2^3+...+2^{2010}\))
=> A= \(2^{2011}-1\)
Mà B = \(2^{2011}\)
=> A < B
A = 2 + 2^2 + 2^3 + 2^4 + ... + 2^2010 hay A = 3 + 2^2 + 2^3 + 2^4 + ... + 2^2010 bạn
i don't now
mong thông cảm !
...........................
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
ta có :
\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{100^2}< \frac{1}{99\cdot100}\)
nên \(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}\)
\(\Rightarrow A< \frac{99}{100}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
nhiều qá lm sao nổi
`Answer:`
\(T=\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2016}{2^{2015}}+\frac{2017}{2^{2016}}\)
\(\Leftrightarrow2T=2+\frac{3}{2}+\frac{4}{2^2}+...+\frac{2016}{2^{2014}}+\frac{2017}{2^{2015}}\)
\(\Leftrightarrow2T-T=2+\left(\frac{3}{2}-\frac{2}{2}\right)+\left(\frac{4}{2^2}-\frac{4}{2^2}\right)+...+\left(\frac{2017}{2^{2015}}-\frac{2016}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)
\(\Leftrightarrow2T-T=2+\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)
Ta đặt \(V=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)
\(\Rightarrow T=2+V-\frac{2017}{2^{2016}}\text{(*)}\)
\(\Leftrightarrow2V=1+\frac{1}{2}+...+\frac{1}{2^{2014}}\)
\(\Leftrightarrow2V-V=\left(1+\frac{1}{2}+...+\frac{1}{2^{2014}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)\)
\(\Leftrightarrow2V-V=1-\frac{1}{2^{2015}}\text{(**)}\)
Từ (*)(**)\(\Rightarrow T=2+\left(1-\frac{1}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)
\(\Leftrightarrow T=3-\frac{1}{2^{2015}}-\frac{2017}{2^{2016}}\)
`=>T<3`
Bài 2 :
\(B=2014\cdot2020\)
\(B=\left(2017-3\right)\left(2017+3\right)\)
\(B=2017^2-3^2\)
\(B=2017^2-9< A=2017^2\)
Vậy \(B< A\)
\(B=2014.2020\)
\(B=\left(2017-3\right)\left(2017+3\right)\)
\(B=\left(2017-3\right).2017+\left(2017+3\right).3\)
\(B=2017^2-3.2017+2017.3+3^2\)
\(B=2017^2-3^2< 2017^2=A\)
Vậy A > B
_Hok tốt_
!!!
a) (-1)7 . 3 với 0
(-1)7 có số mũ lẻ => (-1)7 mang dấu âm => (-1)7 . 3 mang dấu âm
=> (-1)7 . 3 < 0
b) (-1) . 3 . ( -8 ) . 4 . ( -2 ) . ( -5 )2
(-5)2 có số mũ chẵn => (-5)2 mang dấu dương
Nhận thấy có 3 dấu âm , mà lẻ âm thì mang âm => (-1) . 3 . ( -8 ) . 4 . ( -2 ) . ( -5 )2 mang dấu âm
=> (-1) . 3 . ( -8 ) . 4 . ( -2 ) . ( -5 )2 < 0
Có vẻ hơi khó hiểu nhỉ :]
a) Vì 7 là số lẻ \(\Rightarrow\left(-1\right)^7\)là số âm
mà \(3\)dương \(\Rightarrow\left(-1\right)^7.3< 0\)
b) Vì \(\left(-5\right)^2>0\)
mà tích trên có 3 số âm \(\Rightarrow\left(-1\right).3.\left(-8\right).4.\left(-2\right).\left(-5\right)^2< 0\)
\(A=\dfrac{2^{2008}-3}{2^{2007}-1};B=\dfrac{2^{2007}-3}{2^{2006}-1}\)
\(\dfrac{1}{2}A=\dfrac{2^{2008}-3}{2^{2008}-2}=1-\dfrac{1}{2^{2008}-2};\dfrac{1}{2}B=\dfrac{2^{2007}-3}{2^{2007}-2}=1-\dfrac{1}{2^{2007}-2}\)
2^2008-2>2^2007-2
=>1/2^2008-2<1/2^2007-2
=>A>B
Ta có :
\(B=4+2^2+2^3+2^4+...+2^{2016}\)
\(\Rightarrow\) \(B-4=2^2+2^3+2^4+...+2^{2016}\)
\(\Rightarrow\) \(2\left(B-4\right)=2^3+2^4+2^5+...+2^{2017}\)
\(\Rightarrow\) \(2\left(B-4\right)-\left(B-4\right)=B-4=2^{2017}-2^2\)
\(\Rightarrow\) \(B=2^{2017}-2^2+4=2^{2017}\)
\(\Rightarrow\) \(A=B=2^{2017}\)
Vậy \(A=B\)