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1/x+x+1+x+2+x+3+...+x+2006+2007=2007
------------------------------------------=2007-2007
------------------------------------------=0
x+x+x+...+x+1+2+3+...+2006=0
2007.x+(1+2+...+2006)=0
2007.x+(2006+1).[(2006-1)+1]:2=0
2007.x+2013021=0
2007.x=0-2013021
x=-2013021:2007
x=-1003
2/x+x+1+x+2+...+x+198=401-201-200-199
199.x+(1+2+...+198)=-199
199.x+(1+198).[(198-1)+1]:2=-199
199.x+19701=-199
199.x=-199-19701
x=-19900:199
x=-100
3/x+x+1+x+2+...+x+2008=2010-2010-2009
2009.x+(2008+1).[(2008-1)+1]:2=-2009
2009.x+2017036=-2009
2009.x=-2009-2017036
x=-2019045:2009
x=-1005
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Ta có: 1+2-3+...+200-201
Để riêng số 1 ta được:
2-3+3-4+....+200-201 có: 100 cặp
= (-1)+(-1)+...+(-1)
= (-1) x 100 = -100
=> 1+2-3+3-4+...+200-201
= 1 + (-100) = -99
Vậy 1+2-3+...+200-201 = -99
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b)2014/2014*2015=2014:2014/2014*2015:2014=1/2015(rút gọn phân số)
2015/2015*2015=2015:2015/2015*2016:2015=1/2016(rút gọn phân số)
Mà 1/2015>1/2016
=>2014/2014*2015>2015/2015*2015
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(23x5x7)(52x73)/(2x5x72)2=23x(5x52)x(7x73)/22x52x72x2=23x53x74/22x52x74=2x5=10
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a)
Vì \(\frac{2009}{2010}< 1\Rightarrow\frac{2009}{2010}< \frac{2009+1}{2010+1}=\frac{2010}{2011}\)
Cần nhớ:
Nếu: \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+n}{b+n}\left(n\inℕ^∗\right)\)
Và tương tự: \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{b+n}\left(n\inℕ^∗\right)\)
b)Ta có:
\(\frac{1}{3^{400}}=\frac{1}{\left(3^4\right)^{100}}=\frac{1}{81^{100}}\)
\(\frac{1}{4^{300}}=\frac{1}{\left(4^3\right)^{100}}=\frac{1}{64^{100}}\)
Vì: \(81^{100}>64^{100}\Leftrightarrow\frac{1}{81^{100}}< \frac{1}{64^{100}}\Leftrightarrow\frac{1}{3^{400}}< \frac{1}{4^{300}}\)
c) Ta có:
\(\frac{200+201}{201+202}=\frac{401}{403}< 1\)
\(\frac{200}{201}+\frac{201}{202}=1-\frac{1}{201}+1-\frac{1}{202}=2-\left(\frac{1}{201}+\frac{1}{202}\right)>1\)
=>\(\frac{200}{201}+\frac{201}{202}>\frac{200+201}{201+202}\)