Tính S=\(\dfrac{11}{2^2}+\dfrac{11}{3^3}+\dfrac{11}{2^4}+...+\dfrac{11}{2^{2011}}\)
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10 tháng 3 2022
\(2\dfrac{1}{3}.3=\dfrac{7}{3}.3=7.\\ \left(\dfrac{2}{5}-\dfrac{3}{4}\right)-\dfrac{2}{5}=\dfrac{2}{5}-\dfrac{3}{4}-\dfrac{2}{5}=-\dfrac{3}{4}.\\ \dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}.\\ =\dfrac{-10}{11}\left(\dfrac{4}{7}+\dfrac{3}{7}-1\right).\\ =\dfrac{-10}{11}.\left(1-1\right)=0.\)
HN
10 tháng 3 2022
1) 2\(\dfrac{1}{3}\).3=\(\dfrac{7}{3}\).3=7.
2) (2/5 -3/4) -2/5 = 2/5 -3/4 -2/5 = -3/4.
3) \(\dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}=\dfrac{1}{11}\left(-\dfrac{40}{7}-\dfrac{30}{7}+21\right)=\dfrac{1}{11}.\left(-10+21\right)=1\).
Y
15 tháng 8 2023
\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)
Vậy \(A=\dfrac{1}{2}\)
\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)
Vậy \(B=3\)
12 tháng 8 2023
a: =-5/7(2/11+9/11)+12/7
=12/7-5/7
=7/7=1
b: \(=\dfrac{-12}{56}+\dfrac{35}{56}-\dfrac{28}{56}=\dfrac{-5}{56}\)
c: \(=\dfrac{1}{4}-\dfrac{5}{13}+\dfrac{2}{11}-\dfrac{8}{13}+\dfrac{3}{4}\)
=1-1+2/11
=2/11
d: \(=\dfrac{21}{31}+\dfrac{-16}{7}+\dfrac{44}{53}+\dfrac{10}{31}+\dfrac{9}{53}\)
=1+1-16/7
=-2/7
e: \(=\dfrac{\dfrac{4}{36}-\dfrac{30}{36}-\dfrac{144}{36}}{\dfrac{21}{36}-\dfrac{1}{36}-\dfrac{360}{36}}=\dfrac{-160}{-340}=\dfrac{8}{17}\)
17 tháng 4 2023
a. 2/11 + ( -14/11 )
= -12/11
b. 1/4 + ( -3/6 + 4/6 )
= 1/4 + 1/6
= 6/24 + 4/24
= 10/24
= 5/12
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 4 2023
a, \(\dfrac{2}{11}\) + [ \(-\dfrac{5}{11}\) - \(\dfrac{9}{11}\)]
= \(\dfrac{2}{11}\) + [ \(-\dfrac{14}{11}\)]
= \(\dfrac{2-14}{11}\)
= \(-\dfrac{12}{11}\)
b, \(\dfrac{1}{4}\) + ( \(-\dfrac{1}{2}\) + \(\dfrac{2}{3}\))
= \(\dfrac{1}{4}\) + ( \(-\dfrac{3}{6}\) + \(\dfrac{4}{6}\))
= \(\dfrac{1}{4}\) + \(\dfrac{1}{6}\)
= \(\dfrac{3}{12}\) + \(\dfrac{2}{12}\)
= \(\dfrac{5}{12}\)
11 tháng 2 2022
Bài 1 :
\(=\dfrac{2}{11}+\dfrac{4}{11}-\dfrac{6}{11}-\dfrac{3}{8}-\dfrac{5}{8}=0-1=-1\)
Bài 2 :
\(\Rightarrow3+x=8\Leftrightarrow x=5\)
Bài 3 :
\(\Leftrightarrow x-\dfrac{5}{11}=\dfrac{5}{4}\Leftrightarrow x=\dfrac{35}{44}\)
Bài 4 :
Trong 2 ngày An đọc được số quyên phần quyên sách
\(\dfrac{1}{11}+\dfrac{8}{11}=\dfrac{9}{11}\)( quyển sách )
đs : 9/11 quyển sách
MN
17 tháng 7 2021
a) \(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{6}{6}-\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{6-3+2}{6}=\dfrac{1}{6}\)
\(b.\) \(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}=\dfrac{2}{5}+\dfrac{3}{5}.\dfrac{10}{9}=\dfrac{2}{5}+\dfrac{2}{3}=\dfrac{6}{15}+\dfrac{10}{15}=\dfrac{6+10}{15}=\dfrac{16}{15}\)
\(c.\) \(\dfrac{7}{11}.\dfrac{3}{4}+\dfrac{7}{11}.\dfrac{1}{4}+\dfrac{4}{11}=\dfrac{21}{44}+\dfrac{7}{44}+\dfrac{4}{11}=\dfrac{21}{44}+\dfrac{7}{44}+\dfrac{16}{44}=\dfrac{21+7+16}{44}=\dfrac{44}{44}=1\)
17 tháng 7 2021
a/\(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{6}{6}-\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{5}{6}\)
b/\(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}=\dfrac{2}{5}+\dfrac{3}{5}.\dfrac{10}{9}=\dfrac{2}{5}+\dfrac{2}{3}=\dfrac{6}{15}+\dfrac{10}{15}=\dfrac{16}{15}\)
NC
3
29 tháng 5 2022
\(=\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\left(\dfrac{3}{11}-\dfrac{6}{11}\right)+\dfrac{1}{4}=\dfrac{5}{4}-\dfrac{3}{11}=\dfrac{55}{44}-\dfrac{12}{44}=\dfrac{43}{44}\)
KS
29 tháng 5 2022
`2/7-(-3)/11+5/7-1/(-4)-6/11`
`=2/7+3/11+5/7+1/4-6/11`
`=(2/7+5/7)+(3/11-6/11)+1/4`
`=7/7-3/11+1/4`
`=1-3/11+1/4`
`=44/44-12/44+11/44`
`=43/44`
\(S=\dfrac{11}{2^2}+\dfrac{11}{2^3}+\dfrac{11}{2^4}+...+\dfrac{11}{2^{2011}}\)
\(\Rightarrow2S=\dfrac{11}{2}+\dfrac{11}{2^2}+\dfrac{11}{2^3}+...+\dfrac{11}{2^{2010}}\)
\(\Rightarrow2S-S=\left(\dfrac{11}{2}+\dfrac{11}{2^2}+\dfrac{11}{2^3}+...+\dfrac{11}{2^{2010}}\right)\)
\(-\left(\dfrac{11}{2^2}+\dfrac{11}{2^3}+\dfrac{11}{2^4}+...+\dfrac{11}{2^{2011}}\right)\)
\(\Rightarrow S=\dfrac{11}{2}-\dfrac{11}{2^{2010}}=\dfrac{11.2^{2009}}{2^{2010}}-\dfrac{11}{2^{2010}}=\dfrac{11.\left(2^{2009}-1\right)}{2^{2010}}\)
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