1+2+3+ ....+ 2006+2020
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a, Làm
\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x+5}{2016}+\frac{x+6}{2015}\)
<=>\(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}=\frac{x+2021}{2017}+\frac{x+2021}{2016}+\frac{x+2021}{2015}\)
<=>\(\left(x+2021\right)\left(\frac{1}{2020}+\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
<=> x+2021=0
<=> x=-2021
Kl:......................
b, Làmmmmm
\(\frac{2-x}{2004}-1=\frac{1-x}{2005}-\frac{x}{2006}\)
<=> \(\frac{2006-x}{2004}=\frac{2006-x}{2005}+\frac{2006-x}{2006}\)
<=> \(\left(2006-x\right)\left(\frac{1}{2004}-\frac{1}{2005}-\frac{1}{2006}\right)=0< =>2006-x=0\)
<=> x=2006
Kl:..............
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a)\(A=\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{2}\right)vaB=\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
+)Ta có:\(A=\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{2}\right)\)
\(\Leftrightarrow A=\frac{31}{23}-\left(\frac{7}{32}+\frac{128}{32}\right)\)
\(\Leftrightarrow A=\frac{31}{23}-\frac{135}{32}\)
\(\Leftrightarrow A=\frac{992}{736}-\frac{3105}{736}\)
\(\Leftrightarrow A=\frac{-2113}{736}\left(1\right)\)
+)Ta lại có:\(B=\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(\Leftrightarrow B=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(\Leftrightarrow B=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
\(\Leftrightarrow B=\frac{1}{3}+\frac{-67}{67}+\frac{41}{41}\)
\(\Leftrightarrow B=\frac{1}{3}+\left(-1\right)+1\)
\(\Leftrightarrow B=\frac{1}{3}\left(2\right)\)
+)Từ (1) và (2)
\(\Leftrightarrow A< 0< B\Leftrightarrow A< B\)
Vậy A<B
b)\(\frac{200420042004}{200520052005}va\frac{2004}{2005}\)
+)Ta có \(\frac{200420042004}{200520052005}=\frac{2004.100010001}{2005.100010001}=\frac{2004}{2005}\)
\(\Leftrightarrow\frac{200420042004}{200520052005}=\frac{2004}{2005}\)
c)\(C=\frac{2020^{2006}+1}{2020^{2007}+1}vaD=\frac{2020^{2005}+1}{2020^{2006}+1}\)
\(C=\frac{2020^{2006}+1}{2020^{2007}+1}< 1\)
\(\Leftrightarrow C< \frac{2020^{2006}+1+2019}{2020^{2007}+1+2019}=\frac{2020^{2006}+2020}{2020^{2007}+2020}=\frac{2020.\left(2020^{2005}+1\right)}{2020.\left(2020^{2006}+1\right)}=\frac{2020^{2005}+1}{2020^{2006}+1}\)
\(\Leftrightarrow C< D\)
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