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\(n=\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{2005.2010}=\frac{1}{5}\left(\frac{5}{1.5}+\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{2010.2015}\right)\)
\(=\frac{1}{5}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2005}-\frac{1}{2010}\right)=\frac{1}{5}\left(1-\frac{1}{2010}\right)\)
\(=\frac{1}{5}.\frac{2009}{2010}=\frac{2009}{10050}\)
mik nghĩ đây là bài lớp 6
-->n=\(\frac{1}{5}\left(\frac{5}{1.5}+\frac{5}{5.10}+\frac{5}{10.15}+....+\frac{5}{2005.2010}\right)\)
-->n=\(\frac{1}{5}\left[\left(1-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{10}\right)+..+\frac{1}{2005}-\frac{1}{2010}\right]\)
-->n=1/5(1-1/2010)
-->n=2009/2010.1/5
-->n=2009/10050
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li-kecho mk nhé bn
a)5.(-8).2.(-3)
=[5.(-2)].[(-8).(-3)]
=-10.24
=-240
b)(45-135+72)-(45+72)
=45-135+72-45-72
=(45-45)+(72-72)-135
=0+0-135
=-135
c)3.(-5)2.2+2.(-5)-20
=3.25.2+2(-5)-20
=75.2+(-10)-20
=150+(-30)
=120
d)86.(-46)+46.27-46.41
=86.(-1).46+46.27-46.41
=-86.46+46.27-46.41
=46(-86+27-41)
=46(-100)
-4600
e)34(15-10)-15(34-10)
=34.15-34.10-15.34+15.10
=(34.15-15.34)+(15.10-34.10)
=[34(15-15)]+[10(15-34)]
=34.0+10(-9)
=0+(-90)
=-90
\(\frac{1}{5}+\frac{4}{10}+\frac{9}{15}+\frac{16}{20}+\frac{36}{30}+\frac{64}{40}+\frac{81}{45}\)
\(=\frac{33}{5}\)
a) 3/7 + 4/9 + 4/7 + 5/9
= ( 3/7 + 4/7 ) + ( 4/9 + 5/9 )
= 7/7 + 9/9
= 1 + 1
= 2
b)1/5 + 4/10 + 9/15 + 16/20 + 25/25 + 36/30 + 49/35 + 64/40 + 81/45
= 1/5 + 2/5 + 3/5 + 4/5 + 5/5 + 6/5 + 7/5 + 8/5 + 9/5
= ( 1/5 + 9/5 ) + ( 2/5 + 8/5 ) + (7/5 + 3/5 ) + ( 4/5 + 6/5 ) + 5/5
= 2 + 2 + 2 + 2 + 1
= 2 x 4 + 1
= 8 +1
= 9
c) 1/8 + 1/12 + 3/8 + 5/12
= ( 1/8 + 3/8 ) + ( 1/12 + 5/12)
= 4/8 + 6/12
= 1/2 + 1/2
= 2/4 = 1/2
mỏi tay rồi
d; (1 - \(\dfrac{1}{2}\)) x (1 - \(\dfrac{1}{3}\)) x (1 - \(\dfrac{1}{4}\)) x ... x ( 1 - \(\dfrac{1}{100}\))
= \(\dfrac{1}{2}\) x \(\dfrac{2}{3}\) x \(\dfrac{3}{4}\) x \(\dfrac{3}{4}\) x ... x \(\dfrac{99}{100}\)
= \(\dfrac{1}{100}\)
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