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a) Ta có : \(\frac{-3}{100}< 0< \frac{2}{3}\)
\(\Rightarrow\frac{-3}{100}< \frac{2}{3}\)
b) Ta có : \(\frac{267}{268}< 1< \frac{1347}{1343}\)
\(\Rightarrow\frac{267}{268}< \frac{1347}{1343}\)
\(\Rightarrow\frac{267}{-268}< \frac{-1347}{1343}\)
c) Ta có : \(\frac{2017.2018-1}{2017.2018}=\frac{2017.2018}{2017.2018}-\frac{1}{2017.2018}=1-\frac{1}{2017.2018}\)
\(\frac{2018.2019-1}{2018.2019}=\frac{2018.2019}{2018.2019}-\frac{1}{2018.2019}=1-\frac{1}{2018.2019}\)
mà \(2017.2018< 2018.2019\)
\(\Rightarrow\frac{1}{2017.2018}>\frac{1}{2018.2019}\)
\(\Rightarrow1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\)
\(\Rightarrow\frac{2017.2018-1}{2017.2018}< \frac{2018.2019-1}{2018.2019}\)
d) Ta có : \(\frac{2017.2018}{2017.2018+1}=\frac{2017.2018+1}{2017.2018+1}-\frac{1}{2017.2018+1}=1-\frac{1}{2017.2018+1}\)
\(\frac{2018.2019}{2018.2019+1}=\frac{2018.2019+1}{2018.2019+1}-\frac{1}{2018.2019+1}=1-\frac{1}{2018.2019+1}\)
mà \(2017.2018+1< 2018.2019+1\)
\(\Rightarrow\frac{1}{2017.2018+1}>\frac{1}{2018.2019+1}\)
\(\Rightarrow1-\frac{1}{2017.2018+1}< 1-\frac{1}{2018.2019+1}\)
\(\Rightarrow\frac{2017.2018}{2017.2018+1}< \frac{2018.2019}{2018.2019+1}\)
\(B=\frac{1}{1.2}=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(B=\left(1-\frac{1}{2018}\right)-\left(\frac{1}{2}-\frac{1}{2}\right)-...-\left(\frac{1}{2017}-\frac{1}{2017}\right)\)
\(B=1-\frac{1}{2018}=\frac{2017}{2018}\)
Vậy \(B=\frac{2017}{2018}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{102}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{101}{102}=\frac{1}{102}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}=\frac{C}{D}\)
Ta có: \(D=\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}\)(có 2015 số hạng)
\(D=\left(\frac{2015}{1}+1\right)+\left(\frac{2014}{2}+1\right)+...+\left(\frac{1}{2015}+1\right)-2015\)
\(D=2016+\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}-2015\)
\(D=\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}+1=\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}+\frac{2016}{2016}\)
\(D=2016\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right)=2016C\)
Vậy \(B=\frac{C}{D}=\frac{C}{2016C}=\frac{1}{2016}\)
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{102}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{101}{102}=\frac{1\cdot2\cdot3\cdot....\cdot101}{2\cdot3\cdot4\cdot....\cdot102}\)
\(A=\frac{1}{102}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\left(\frac{2015}{1}+1\right)+\left(\frac{2014}{2}+1\right)+...+\left(\frac{1}{2015}+1\right)+1}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\frac{2016}{1}+\frac{2016}{2}+...+\frac{2016}{2015}+\frac{2016}{2016}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{2016\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}=\frac{1}{2016}\)
1. a) \(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{1}{2}+\frac{1}{3}=\frac{9}{12}+\frac{6}{12}+\frac{4}{12}=\frac{19}{12}\)
b) \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}\)
\(=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}\)
\(=5+1+0,5=6,5\)
2) a) 1/2 + 2/3x = 1/4
=> 2/3x = 1/4 - 1/2
=> 2/3x = -1/4
=> x = -1/4 : 2/3
=> x = -3/8
b) 3/5 + 2/5 : x = 3 1/2
=> 3/5 + 2/5 : x = 7/2
=> 2/5 : x = 7/2 - 3/5
=> 2/5 : x = 29/10
=> x = 2/5 : 29/10
=> x = 4/29
c) x+4/2004 + x+3/2005 = x+2/2006 + x+1/2007
=> x+4/2004 + 1 + x+3/2005 + 1 = x+2/2006 + 1 + x+1/2007 + 1
=> x+2008/2004 + x+2008/2005 = x+2008/2006 + x+2008/2007
=> x+2008/2004 + x+2008/2005 - x+2008/2006 - x+2008/2007 = 0
=> (x+2008). (1/2004 + 1/2005 - 1/2006 - 1/2007) = 0
Vì 1/2004 + 1/2005 - 1/2006 - 1/2007 khác 0
Nên x + 2008 = 0 <=> x = -2008
Vậy x = -2008
1,a,\(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{2}{4}+\frac{1}{3}=\frac{5}{4}+\frac{1}{3}=\frac{15}{12}+\frac{4}{12}=\frac{19}{12}\)
b, \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}=5+1+\frac{1}{2}=\frac{13}{2}\)2,a,\(\frac{1}{2}+\frac{2}{3}.x=\frac{1}{4}\)
<=>\(\frac{2}{3}.x=-\frac{1}{2}\)
<=>\(x=-\frac{3}{4}\)
b,\(\frac{3}{5}+\frac{2}{5}\div x=3\frac{1}{2}\)
<=>\(\frac{2}{5x}=\frac{29}{10}\)
<=>\(x=\frac{29}{4}\)
c,\(\frac{x+4}{2004}+\frac{x+3}{2005}=\frac{x+2}{2006}+\frac{x+1}{2007}\)
<=> \(\frac{x+4}{2004}+1+\frac{x+3}{2005}+1=\frac{x+2}{2006}+1+\frac{x+1}{2007}+1\)
<=>\(\frac{x+2008}{2004}+\frac{x+2008}{2005}=\frac{x+2008}{2006}+\frac{x+2008}{2007}\)
<=>\(\left(x+2008\right)\left(\frac{1}{2004}+\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)\)=0
<=>x+2008=0 vì cái ngoặc còn lại\(\ne0\)
<=>x=-2008
Vậy x=-2008
Bạn nhớ tk cho mình vì mình đã chăm chỉ làm hết bài bạn hỏi nha!
\(taco:A=\frac{1}{2\cdot32}+\frac{1}{3\cdot33}+\frac{1}{4\cdot34}+...+\frac{1}{1988\cdot2018}\)
\(A=\frac{1}{2}-\frac{1}{32}+\frac{1}{3}-\frac{1}{33}+\frac{1}{4}-\frac{1}{34}+...+\frac{1}{1988}-\frac{1}{2018}\)
\(A=\frac{1}{2}-\frac{1}{2018}\)
\(A=\frac{504}{1009}\)
\(B=\frac{1}{2\cdot1989}+\frac{1}{3\cdot1990}+\frac{1}{4\cdot1991}+...+\frac{1}{31\cdot2018}\)
\(B=\frac{1}{2}-\frac{1}{1989}+\frac{1}{3}-\frac{1}{1990}+\frac{1}{4}-\frac{1}{1991}+...+\frac{1}{31}-\frac{1}{2018}\)
\(B=\frac{1}{2}-\frac{1}{1018}\)
\(B=\frac{504}{1009}\)
tu do \(=>\frac{504}{\frac{1009}{\frac{504}{1009}}}< =>1\)
chuc ban hoc tot nhe >.<