C = \(\dfrac{\dfrac{1}{285}+\dfrac{1}{286}+\dfrac{1}{287}+...+\dfrac{1}{568}}{\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{567}-\dfrac{1}{568}}\) . Tính C
Giải chi tiết hộ nha__camon
98775 - 32 x 85 / 67500 - 24 x 236 / 568 + 101598 : 287 / 6875 + 980 -180 \(\dfrac{2}{5}+\dfrac{3}{10}-\dfrac{1}{2}\) / \(\dfrac{8}{11}+\dfrac{8}{33}\) x \(\dfrac{3}{4}\) / 7/9 x 3/14 :5/8 / \(\dfrac{5}{12}-\dfrac{7}{32}:\dfrac{21}{16}\)
ÉT O ÉT
98775 - 32 x 85
=98775 -2720
=96055
67500 - 24 x 236
= 67500 -5664
=61836
568 + 101598 : 287
= 568 +354
=922
6875 + 980 -180
=7855 -180
=7675
\(\dfrac{2}{5}+\dfrac{3}{10}-\dfrac{1}{2}\)
\(=\dfrac{7}{10}-\dfrac{1}{2}\)
= \(\dfrac{1}{5}\)
\(\dfrac{8}{11}+\dfrac{8}{33}x\dfrac{3}{4}\)
\(=\dfrac{8}{11}+\dfrac{2}{11}\)
\(=\dfrac{10}{11}\)
\(\dfrac{7}{9}x\dfrac{3}{14}:\dfrac{5}{8}\)
\(=\dfrac{1}{6}:\dfrac{5}{8}\)
\(=\dfrac{1}{6}x\dfrac{8}{5}\)
\(=\dfrac{8}{30}\)
\(=\dfrac{4}{15}\)
\(\dfrac{5}{12}-\dfrac{7}{32}:\dfrac{21}{16}\)
\(=\dfrac{5}{12}-\dfrac{7}{32}x\dfrac{16}{21}\)
\(=\dfrac{5}{12}-\dfrac{1}{6}\)
\(=\dfrac{5}{12}-\dfrac{2}{12}\)
\(=\dfrac{3}{12}=\dfrac{1}{4}\)
1) So sánh
a/ -3\(\dfrac{1}{5}\) ; \(\dfrac{\text{37}}{10}\) và -3,25 (đổi ra số thập phân)
b/\(\dfrac{-567}{568}\) và \(\dfrac{-568}{567}\) (so sánh với số 1)
c/ -0,625 và \(\dfrac{-5}{7}\) (đổi ra phân số)
2) Tính
a/ 3,6 - \(\dfrac{-5}{6}\) . (- 2,4) . \(\dfrac{3}{5}\)
b/ \(\dfrac{1}{4}\) - (0,5) - 6\(\dfrac{1}{2}\) + \(\dfrac{5}{8}\)
c/ 1,1 - ( - 1,2 ) -\(\left|-1,3\right|\) - 2\(\dfrac{3}{4}\)
Giúp mình với nha
mình gần nộp rồi
mình hứa sẽ tick
Bài 1:
a) Ta có:
\(-3\dfrac{1}{5}=-2,8\)
\(\dfrac{37}{10}=3,7\)
Vì \(-3,25< -2,8< 3,7\)
Nên \(-3,25< -3\dfrac{1}{5}< \dfrac{37}{10}\)
b) Ta có:
\(\dfrac{-567}{568}>-1\)
\(\dfrac{-568}{567}< -1\)
\(\Leftrightarrow\dfrac{-567}{568}>-1>\dfrac{-568}{567}\)
\(\Leftrightarrow\dfrac{-567}{568}>\dfrac{-568}{567}\)
c) \(-0,625=-\dfrac{5}{8}\)
Vì \(8>7\)
\(\Leftrightarrow\dfrac{1}{8}< \dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{-5}{8}>\dfrac{-5}{7}\)
\(\Leftrightarrow-0,625>\dfrac{-5}{7}\)
Vậy ...
Giải:
a) \(3,6-\dfrac{-5}{6}.\left(-2,4\right).\dfrac{3}{5}\)
\(=\dfrac{18}{5}-\dfrac{-5}{6}.\left(-\dfrac{12}{5}\right).\dfrac{3}{5}\)
\(=\dfrac{18}{5}-\dfrac{6}{5}\)
\(=\dfrac{12}{5}\)
Vậy ...
b) \(\dfrac{1}{4}-0,5-6\dfrac{1}{2}+\dfrac{5}{8}\)
\(=\dfrac{1}{4}-\dfrac{1}{2}-6\dfrac{1}{2}+\dfrac{5}{8}\)
\(=-\dfrac{49}{8}\)
Vậy ...
c) \(1,1-\left(-1,2\right)-\left|-1,3\right|-2\dfrac{3}{4}\)
\(=1,1+1,2-1,3-2,75\)
\(=-\dfrac{7}{4}=-1,75\)
Vậy ...
Tính: \(E=\dfrac{\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right)...\left(\dfrac{1}{2002}-1\right).\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{9999}{10000}}\)
Giải chi tiết giúp mình nha. Thanks
\(E=\dfrac{\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2002}-1\right)\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{9999}{10000}}\)
\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{100^2}\right)}\)
\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{100}{101}\cdot\dfrac{101}{102}\cdot...\cdot\dfrac{2002}{2003}}{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)}\)
\(=\dfrac{100}{2003}:\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\right)\)
\(=\dfrac{100}{2003}:\left(\dfrac{101}{2}\right)=\dfrac{100}{2003}\cdot\dfrac{2}{101}=\dfrac{200}{202303}\)
Tính hợp lí:
a) A = \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\) + \(\dfrac{1}{90}\) + \(\dfrac{1}{110}\) + \(\dfrac{1}{132}\) + \(\dfrac{1}{156}\) ;
b) B = \(\dfrac{4}{21}\) + \(\dfrac{4}{77}\) + \(\dfrac{4}{165}\) + \(\dfrac{4}{285}\) +\(\dfrac{4}{437}\) +\(\dfrac{4}{621}\);
c) C = \(\dfrac{1}{21}\) + \(\dfrac{1}{77}\) +\(\dfrac{1}{165}\) +\(\dfrac{1}{285}\) +\(\dfrac{1}{437}\) +\(\dfrac{1}{621}\) ;
d) D = \(\dfrac{1}{1.6}\) + \(\dfrac{1}{6.11}\) +\(\dfrac{1}{11.16}\) +\(\dfrac{1}{16.21}\) +\(\dfrac{1}{26.31}\) .
Giải:
a)A=1/56+1/72+1/90+1/110+1/132+1/156
A=1/7.8+1/8.9+1/9.10+1/10.11+1/11.12+1/12.13
A=1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11+1/11-1/12+1/12-1/13
A=1/7-1/13
A=6/91
b)B=4/21+4/77+4/165+4/285+4/437+4/621
B=4/3.7+4/7.11+4/11.15+4/15.19+4/19.23+4/23.27
B=1/3-1/7+1/7-1/11+1/11-1/15+1/15-1/19+1/19-1/23+1/23-1/27
B=1/3-1/27
B=8/27
c) C=1/21+1/77+1/165+1/285+1/437+1/621
C=1/3.7+1/7.11+1/11.15+1/15.19+1/19.23+1/23.27
C=1/4.(4/3.7+4/7.11+4/11.15+4/15.19+4/19.23+4/23.27)
C=1/4.(1/3-1/7+1/7-1/11+1/11-1/15+1/15-1/19+1/19-1/23+1/23-1/27)
C=1/4.(1/3-1/27)
C=1/4.8/27
C=2/27
d) D=1/1.6+1/6.11+1/11.16+1/16.21+1/21.26+1/26.31
D=1/5.(5/1.6+5/6.11+5/11.16+5/16.21+5/21.26+5/26.31)
D=1/5.(1/1-1/6+1/6-1/11+1/11-1/16+1/16-1/21+1/21-1/26+1/26-1/31)
D=1/5.(1/1-1/31)
D=1/5.30/31
D=6/31
Nếu câu d cậu viết thiếu thì làm như vầy nhé!
Chúc bạn học tốt!
Nếu như câu d ko chép sai thì làm thế này nha:
d) D=1/1.6+1/6.11+1/11.16+1/16.21+1/26.31
D=1/5.(5/1.6+5/6.11+5/11.16+5/16.21)+1/806
D=1/5.(1/1-1/6+1/6-1/11+1/11-1/16+1/16-1/21)+1/806
D=1/5.(1/1-1/21)+1/806
D=1/5.20/21+1/806
D=4/21+1/806
D=3245/16926
Chúc bạn học tốt!
Hãy tính \(\dfrac{C}{D}\). Biết C= \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\) và D= \(\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
=> D + 49 = (1/49 + 1) + (2/48 + 1) +... (49/1 + 1)
= 50/1 + 50/2 + ... + 50/49
= 50(1/2+1/3+...+1/49) + 50
=> D = 50(1/2 + 1/3 +... + 1/49) + 1
= 50(1/2 + 1/3 +... + 1/49 + 1/50)
=> C/D = 1/50
Tìm \(lim\) \(u_n\), biết \(u_n=\dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+...+\dfrac{1}{n^2-1}\).
A. \(lim\) \(u_n=\dfrac{3}{4}\).
B. \(lim\) \(u_n=\dfrac{3}{5}\).
C. \(lim\) \(u_n=\dfrac{2}{3}\).
D. \(lim\) \(u_n=\dfrac{4}{3}\).
Giải thích chi tiết bước làm và tại sao lại làm như vậy.
\(u_n=\dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+...+\dfrac{1}{n^2-1}\)
\(=\dfrac{1}{\left(2-1\right)\left(2+1\right)}+\dfrac{1}{\left(3-1\right)\left(3+1\right)}+...+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)
\(=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+...+\dfrac{1}{\left(n-1\right)\cdot\left(n+1\right)}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+...+\dfrac{2}{\left(n-1\right)\left(n+1\right)}\right)\)
\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+...+\dfrac{1}{\left(n-1\right)}-\dfrac{1}{\left(n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n+1}\right)=\dfrac{1}{2}\cdot\left(\dfrac{3}{2}-\dfrac{1}{n+1}\right)\)
\(=\dfrac{3}{4}-\dfrac{1}{2n+2}\)
\(\lim\limits u_n=\lim\limits\left(\dfrac{3}{4}-\dfrac{1}{2n+2}\right)\)
\(=\lim\limits\dfrac{3}{4}-\lim\limits\dfrac{1}{2n+2}\)
\(=\dfrac{3}{4}-\lim\limits\dfrac{\dfrac{1}{n}}{2+\dfrac{1}{n}}\)
=3/4
=>Chọn A
4. Tìm x biết:
\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)
Giải chi tiết giúp mình nha.
\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\)- \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\) - \(\dfrac{1}{10.11}\) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+ \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\) + \(\dfrac{1}{10.11}\) =\(x\)-\(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{11}\) = \(x\) - \(\dfrac{5}{13}\)
\(x-\dfrac{5}{13}=\dfrac{1}{11}\)
\(x\) = \(\dfrac{1}{11}\) + \(\dfrac{5}{13}\)
\(x\) = \(\dfrac{68}{143}\)
Tính (theo mẫu).
Mẫu: \(2+\dfrac{1}{6}=\dfrac{12}{6}+\dfrac{1}{6}=\dfrac{13}{6};1-\dfrac{1}{4}=\dfrac{4}{4}-\dfrac{1}{4}=\dfrac{3}{4}\) |
a) \(1+\dfrac{4}{9}\) b) \(5+\dfrac{1}{2}\) c) \(3-\dfrac{5}{6}\) d) \(\dfrac{31}{7}-2\)
a) \(1+\dfrac{4}{9}=\dfrac{9}{9}+\dfrac{4}{9}=\dfrac{9+4}{9}=\dfrac{13}{9}\)
b) \(5+\dfrac{1}{2}=\dfrac{10}{2}+\dfrac{1}{2}=\dfrac{10+1}{2}=\dfrac{11}{2}\)
c) \(3-\dfrac{5}{6}=\dfrac{18}{6}-\dfrac{5}{6}=\dfrac{18-5}{6}=\dfrac{13}{6}\)
d) \(\dfrac{31}{7}-2=\dfrac{31}{7}-\dfrac{14}{7}=\dfrac{31-14}{7}=\dfrac{17}{7}\)
a)(\(\dfrac{31}{35}\)-\(\dfrac{4}{7}\)) x \(\dfrac{8}{7}\) : 2
b)(1 - \(\dfrac{1}{2}\)) x ( 1- \(\dfrac{1}{3}\)) x (1-\(\dfrac{1}{4}\)) x ( 1-\(\dfrac{1}{5}\))
nhanh giúp mik với!!
lời giải chi tiết ạ!!!
\(a,\left(\dfrac{31}{35}-\dfrac{4}{7}\right)\times\dfrac{8}{7}:2\\ =\left(\dfrac{31}{35}-\dfrac{4\times5}{35}\right)\times\dfrac{8}{7}:2\\ =\dfrac{11}{35}\times\dfrac{8}{7}:2\\ =\dfrac{88}{245}:2\\ =\dfrac{44}{245}\\ b,\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\\ =\left(\dfrac{2-1}{2}\right)\times\left(\dfrac{3-1}{3}\right)\times\left(\dfrac{4-1}{4}\right)\times\left(\dfrac{5-1}{5}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)
a, ( \(\dfrac{31}{35}\) - \(\dfrac{4}{7}\)) \(\times\) \(\dfrac{8}{7}\): 2
= \(\left(\dfrac{31}{35}-\dfrac{20}{35}\right)\) \(\times\) \(\dfrac{8}{7}\) : 2
= \(\dfrac{11}{35}\) \(\times\) \(\dfrac{8}{7}\) \(\times\) \(\dfrac{1}{2}\)
= \(\dfrac{44}{35}\) \(\times\) \(\dfrac{4}{7}\)
= \(\dfrac{44}{245}\)
b, ( 1 - \(\dfrac{1}{2}\)) \(\times\) ( 1 - \(\dfrac{1}{3}\)) \(\times\) ( 1 - \(\dfrac{1}{4}\)) \(\times\) ( 1 - \(\dfrac{1}{5}\))
= \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{4}\) \(\times\) \(\dfrac{4}{5}\)
= \(\dfrac{1}{5}\) \(\times\) \(\dfrac{2\times3\times4}{2\times3\times4}\)
= \(\dfrac{1}{5}\)