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vì 1/9 > 1/40 ; 1/29 > 1/40 ; 1/31 > 1/40; 1/39 > 1/40
nên 1/9 + 1/ 29 + 1/31 + 1/39 > 1/40 + 1/40 + 1/40 + 1/40 mà 1/40 + 1/40 + 1/40 + 1/40 = 1/10
=) M > 1/10
M > 1/20 + 1/30 + 1/40 + 1/40
M> 2/15 > 2/20 = 1/10
=> M > 1/10
e: \(\dfrac{2}{5}=\dfrac{8}{20}\)
\(\dfrac{3}{4}=\dfrac{15}{20}\)
mà 8<15
nên \(\dfrac{2}{5}< \dfrac{3}{4}\)
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2009\cdot2011}\)
\(=\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{2011-2009}{2009\cdot2011}\)
\(=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{2009\cdot2011}\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{2011}\right)\)
\(=\frac{1}{2}\cdot\frac{2010}{2011}=\frac{1005}{2011}\)
Đặt \(A=\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{2009\times2011}\)
\(2A=\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{2009\times2011}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\)
\(2A=1-\frac{1}{2011}=\frac{2010}{2011}\Rightarrow A=\frac{1005}{2011}\)
a: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)
\(=\dfrac{35}{6}:\dfrac{95-126}{30}-\dfrac{11}{31}\)
\(=\dfrac{35}{6}\cdot\dfrac{30}{-31}-\dfrac{11}{31}\)
\(=\dfrac{-35\cdot5}{31}-\dfrac{11}{31}=\dfrac{-186}{31}=-6\)
b: \(=\left(-8\right)\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=-4:\dfrac{27-14}{12}=\dfrac{-4\cdot12}{13}=\dfrac{-48}{13}\)
5.42 + 32.5.2 - 1
= 5.16 + 9.(5.2) - 1
= 80 + 9.10 - 1
= 80 + 90 - 1
= 170 - 1
= 169
5 . 42 + 32 . 5 . 2 -1
= 5 . 16 + 9 . 5 . 2 - 1
= 80 + 45 . 2 - 1
= 80 + 90 - 1
= 170 - 1
= 169