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Đặt D1 = \(\dfrac{2010}{1}\) + \(\dfrac{2009}{2}\) + \(\dfrac{2008}{3}\) + ... + \(\dfrac{1}{2010}\)
= 1 + ( 1+ \(\dfrac{2009}{2}\)) + ( 1+ \(\dfrac{2008}{3}\)) + ... + (1+\(\dfrac{1}{2010}\))
= \(\dfrac{2011}{2}\) + \(\dfrac{2011}{3}\)+ ... + \(\dfrac{2011}{2010}\) + \(\dfrac{2011}{2011}\)
= 2011. ( \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2010}\) + \(\dfrac{1}{2011}\))
Đặt D2 = \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2010}\) + \(\dfrac{1}{2011}\)
=> D = 2011
cho mk 1 tick nha 
3.
\(A=1-3+3^2-3^3+...-3^{2009}-3^{2010}\)
\(\Rightarrow3A=3-3^2+3^3-3^4+...-3^{2010}+3^{2011}\)
\(\Rightarrow4A=3-3^2+3^3-3^4+...-3^{2010}+3^{2011}+\left(1-3+3^2-3^3+...-3^{2009}+3^{2010}\right)\)\(\Rightarrow4A=3^{2011}-1\)
\(\Rightarrow4A=3^{2011}\)
\(\Rightarrow\)ĐPCM
mik đg cần gấp câu 1 và câu 2 câu 3 mik lm đk r mơn bn nhé nhưng bn có thể giúp mik câu 1 và câu 2 k?
\(M=\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+59}\\ =\dfrac{1}{\dfrac{3\cdot4}{2}}+\dfrac{1}{\dfrac{4\cdot5}{2}}+...+\dfrac{1}{\dfrac{59\cdot60}{2}}\\ =\dfrac{2}{3\cdot4}+\dfrac{2}{4\cdot5}+...+\dfrac{2}{59\cdot60}\\ =2\left(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{59\cdot60}\right)\\ =2\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{59}-\dfrac{1}{60}\right)\\ =2\cdot\dfrac{19}{60}\\ =\dfrac{38}{60}< \dfrac{40}{60}=\dfrac{2}{3}\)
2, ta thấy:
\(\dfrac{2008}{2009}< \dfrac{2008}{2009+2010}\left(1\right)\)
\(\dfrac{2009}{2010}< \dfrac{2009}{2009+20010}\left(2\right)\)
từ (1) và (2) cộng vế với vế ta đc :\(\dfrac{2008}{2009}+\dfrac{2009}{20010}< \dfrac{2008}{2009+2010}+\dfrac{2009}{2009+2010}=\dfrac{2008+2009}{2009+2010}\)
\(Q=\dfrac{1}{2011}+\dfrac{2}{2010}+\dfrac{3}{2009}+...+\dfrac{2010}{2}+\dfrac{2011}{1}\)
\(Q=\left(1+\dfrac{2}{2011}\right)\left(1+\dfrac{2}{2010}\right)+\left(1+\dfrac{3}{2009}\right)+...+\left(1+\dfrac{2010}{2}\right)+1\)
\(Q=\dfrac{2012}{2011}+\dfrac{2012}{2010}+\dfrac{2012}{2009}+...+\dfrac{2012}{2}+\dfrac{2012}{2012}\)
\(Q=2012.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\)
\(\Rightarrow\dfrac{P}{Q}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}}{2012.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)}=\dfrac{1}{2012}\)
a) M =1+3+32+33+......+3118+3119
M = ( 1+3+32 ) +...+ ( 3117 + 3118+3119 )
M = 1. ( 1+3+32 ) + ... + 3117 . ( 3117 + 3118+3119 )
M = ( 1+3+32 ) .( 1 + ... + 3117 )
M = 13 . ( 1 + ... + 3117 ) \(⋮\) 13 (đpcm )
b) Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...
\(\dfrac{1}{2009^2}< \dfrac{1}{2008.2009}\)
\(\dfrac{1}{2010^2}< \dfrac{1}{2009.2010}\)
=> \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2009^2}+\dfrac{1}{2010^2}\) < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}+\dfrac{1}{2009.2010}\) (1)
Biến đổi vế trái:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}+\dfrac{1}{2009.2010}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2009}-\dfrac{1}{2010}\)
= \(1-\dfrac{1}{2010}\)
= \(\dfrac{2009}{2010}< 1\) (2)
Từ (1) và (2), suy ra :
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2009^2}+\dfrac{1}{2010^2}\) < 1 hay:
N < 1
\(\dfrac{x+1}{2011}+\dfrac{x+2}{2010}+\dfrac{x+3}{2009}+\dfrac{x+4}{2008}=-4\)
\(\Rightarrow\dfrac{x+1}{2011}+1+\dfrac{x+2}{2010}+1+\dfrac{x+3}{2009}+1+\dfrac{x+4}{2008}+1=0\)
\(\Rightarrow\dfrac{x+2012}{2011}+\dfrac{x+2012}{2010}+\dfrac{x+2012}{2009}+\dfrac{x+2012}{2008}=0\)
\(\Rightarrow\left(x+2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}\right)=0\)
Mà \(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}\ne0\)
\(\Rightarrow x+2012=0\Rightarrow x=-2012\)
Vậy x = -2012
tính M hay chứng minh M ko là stn hay đầu bài là j vậy bn????
chek la tinh M day ban