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a) Ta có : \(\dfrac{-1}{5}< 0< \dfrac{1}{1000}\)
\(\Rightarrow\dfrac{-1}{5}< \dfrac{1}{1000}\)
b) Ta có : \(\dfrac{267}{268}< 1< \dfrac{1347}{1343}\)
=> \(\dfrac{267}{-268}< -\dfrac{1347}{1343}\)
c) \(\dfrac{13}{38}>\dfrac{13}{39}=\dfrac{1}{3}=\dfrac{19}{87}>\dfrac{29}{88}\)
=> \(-\dfrac{13}{38}< \dfrac{29}{-88}\)
d) \(\dfrac{181818}{313131}=\dfrac{18}{31}\)
=> \(-\dfrac{18}{31}=-\dfrac{181818}{313131}\)
a)Ta có :
\(-\dfrac{265}{317}< -\dfrac{83}{317}< -\dfrac{83}{111}\Rightarrow-\dfrac{265}{317}< -\dfrac{83}{111}\)
b)Ta có :
\(\dfrac{2002}{2003}< 1< \dfrac{14}{13}\Rightarrow\dfrac{2002}{2003}< \dfrac{14}{13}\)
c)Ta có :
\(\dfrac{-1}{-3}=\dfrac{1}{3}\Rightarrow-\dfrac{27}{463}< 0< \dfrac{1}{3}\Rightarrow-\dfrac{27}{463}< \dfrac{1}{3}\)
Chữa lại đề.Bạn xem lại đề xem đúng chưa nhé!
\(D=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}+\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}+\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}+\dfrac{3}{2004}}\)
\(D=\dfrac{1.\left(\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}\right)}{5.\left(\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}\right)}-\dfrac{2.\left(\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)}{3\left(\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)}\)
\(D=\dfrac{1}{5}-\dfrac{2}{3}\)
\(D=-\dfrac{7}{15}\)
Cái này học lâu rồi.Bạn xem lại xem mình làm đúng chưa nhé!
a, \(\left(2-\dfrac{3}{2}\right)\left(2-\dfrac{4}{3}\right)\left(2-\dfrac{5}{4}\right)\left(2-\dfrac{6}{5}\right)\)
\(=\left(\dfrac{4}{2}-\dfrac{3}{2}\right)\left(\dfrac{6}{3}-\dfrac{4}{3}\right)\left(\dfrac{8}{4}-\dfrac{5}{4}\right)\left(\dfrac{10}{5}-\dfrac{6}{5}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\)
\(=\dfrac{1}{5}\)
b. \(\dfrac{1}{2002}+\dfrac{2003.2001}{2002}-2003\)\(=\dfrac{1}{2002}+\dfrac{2003.2001}{2002}-\dfrac{2003.2002}{2002}\) = \(\dfrac{1+2003.2001-2003.2002}{2002}\) = \(\dfrac{1+\left(2003\left(2001-2002\right)\right)}{2002}\) = \(\dfrac{1+2003.\left(-1\right)}{2002}\) = \(\dfrac{1+\left(-2003\right)}{2002}\) = \(\dfrac{-2002}{2002}=-1\)
Chúc nguyễn hồng nhung học tốt
6:
\(4D=2^2+2^4+...+2^{202}\)
=>3D=2^202-1
hay \(D=\dfrac{2^{202}-1}{3}\)
7: \(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{32}{99}=\dfrac{16}{99}\)
bài 1\(\dfrac{1}{2002}+\dfrac{2003\cdot2001}{2002}+2003=\dfrac{1+2003\cdot2001+2003\cdot2002}{2002}=\dfrac{1+2003\left(2001+2003\right)}{2002}=1+2003\cdot2=4007\)
câu3
a)VP=\(\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{a+1-a}{a\left(a+1\right)}=\dfrac{1}{a\left(a+1\right)}\)=VT
b)VP=VT\(\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}=\dfrac{a+2}{a\left(a+1\right)\left(a+2\right)}-\dfrac{a}{a\left(a+1\right)\left(a+2\right)}=\dfrac{a+2-a}{a\left(a+1\right)\left(a+2\right)}=\dfrac{2}{a\left(a+1\right)\left(a+2\right)}\)
-1/540<0<1/3780
\(\dfrac{-2003}{2004}>-1>-\dfrac{2005}{2003}\)