\(1\) chia \(\dfrac{1}{40}\)
Bài 1:
a) Tìm số tự nhiên nhỏ nhất sao cho khi chia 11 dư 6,chia 4 dư 1 và chia cho 19 dư 11.
b) Cho B=1.2.3........2012.(1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+...+\(\dfrac{1}{2012}\))
Bài 2:
a) Cho a,b là các số nguyên tố thỏa mãn (a2+b2) chia hết cho 3.Chứng minh rằng a và b đều chia hết cho 3
b)Tìm 2 số nguyên tố x và y sao cho x2-6y2=1
Bài 3:So sánh
P=\(\dfrac{2010}{2011}\)+\(\dfrac{2011}{2012}\)+\(\dfrac{2012}{2013}\)và Q=\(\dfrac{2010+2011+2012}{2011+2012+2013}\)
Mọi người giúp mình vs . Mình cảm ơn mọi người nhiều nha!!!
A=(\(\dfrac{1}{8.14}+\dfrac{1}{14.20}+\dfrac{1}{20.26}+...+\dfrac{1}{50.56}\))
B=(\(\dfrac{45}{12.21}+\dfrac{45}{21.30}-\dfrac{40}{24.30}-\dfrac{40}{34.44}-\dfrac{40}{44.54}-\dfrac{40}{54.64}\))
Chứng minh:
\(\dfrac{A}{B}< \dfrac{1}{8}\)
Giúp mình với mình sắp kiểm tra rồi!!
\(A=\dfrac{1}{8.14}+\dfrac{1}{14.20}+\dfrac{1}{20.26}+...+\dfrac{1}{50.56}\)
\(A=\dfrac{1}{6}.\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+...+\dfrac{6}{50.56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+...+\dfrac{1}{50}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{1}{8}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{7}{56}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\dfrac{6}{56}\)
\(A=\dfrac{1}{1}.\dfrac{1}{56}\)
\(A=\dfrac{1}{56}\)
\(B=\dfrac{45}{12.21}+\dfrac{45}{21.30}-\dfrac{40}{24.34}-\dfrac{40}{34.44}-\dfrac{40}{44.54}-\dfrac{40}{54.64}\)
\(B=5\left(\dfrac{9}{12.21}+\dfrac{9}{21.30}\right)-4\left(\dfrac{10}{24.34}+\dfrac{10}{34.44}+\dfrac{10}{44.54}+\dfrac{10}{54.64}\right)\)
\(B=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{54}+\dfrac{1}{54}-\dfrac{1}{64}\right)\)\(B=5\left(\dfrac{5}{60}-\dfrac{2}{60}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{64}\right)\)
\(B=5.\dfrac{3}{60}-\left(\dfrac{4}{24}-\dfrac{4}{64}\right)\)
\(B=5.\dfrac{1}{20}-\left(\dfrac{1}{6}-\dfrac{1}{16}\right)\)
\(B=\dfrac{5}{20}-\left(\dfrac{8}{48}-\dfrac{3}{48}\right)\)
\(B=\dfrac{1}{4}-\dfrac{5}{48}\)
\(B=\dfrac{12}{48}-\dfrac{5}{48}\)
\(B=\dfrac{7}{48}\)
\(\dfrac{A}{B}=\dfrac{1}{56}:\dfrac{7}{48}\)
\(\dfrac{A}{B}=\dfrac{1}{56}.\dfrac{48}{7}\)
\(\dfrac{A}{B}=\dfrac{1}{7}.\dfrac{6}{7}\)
\(\dfrac{A}{B}=\dfrac{6}{49}=\dfrac{48}{392}< \dfrac{49}{392}=\dfrac{1}{8}\)
\(\dfrac{A}{B}< \dfrac{1}{8}\)
Vậy \(\dfrac{A}{B}< \dfrac{1}{8}\)
\(\dfrac{1}{3}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{40}\) + ... + \(\dfrac{1}{101,103}\)
Tính
a) A = \(\dfrac{1}{2.9}+\dfrac{1}{9.7}+\dfrac{1}{7.19}+...+\dfrac{1}{252.509}\)
b) B =\(\dfrac{1}{10.9}+\dfrac{1}{18.13}+\dfrac{1}{26.17}+...+\dfrac{1}{802.405}\)
c) C = \(\dfrac{2}{4.7}-\dfrac{3}{5.9}+\dfrac{2}{7.10}-\dfrac{3}{9.13}+...+\dfrac{2}{301.304}-\dfrac{3}{401.405}\)
giúp mk với
b)\(\dfrac{1}{7}B=\dfrac{1}{10.18}+\dfrac{1}{18.26}+\dfrac{1}{26.34}+...+\dfrac{1}{802.810}\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{8}{10.18}+\dfrac{8}{18.26}+\dfrac{8}{26.34}+...+\dfrac{8}{802.810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{1}{10}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{34}+...+\dfrac{1}{802}-\dfrac{1}{810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{1}{10}-\dfrac{1}{810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}.\dfrac{8}{81}\)
\(\dfrac{1}{7}B=\dfrac{1.8}{8.81}\)
\(\dfrac{1}{7}B=\dfrac{1}{81}\)
\(B=\dfrac{1}{81}:\dfrac{1}{7}\)
\(B=\dfrac{7}{81}\)
Tìm x:
a)\(2016x+\left(\dfrac{7}{12}+\dfrac{4}{21}+\dfrac{2}{24}+\dfrac{11}{30}+\dfrac{3}{40}+\dfrac{15}{56}\right)-\left(\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}\right)=0\)
b\(\dfrac{2x-1}{x+2015}-\dfrac{4025}{x+2017}=\dfrac{x-2014}{2x-4036}-\dfrac{x-2013}{2x-4030}\) (x thuộc N)
c)\(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{x\left(x+2\right)}\right)=\dfrac{4016}{2007}\)
AI GIÚP MK VỚI MK TICK CHO
Tính \(\dfrac{A}{B}\) biết:
\(A=\dfrac{1}{2.17}+\dfrac{1}{3.18}+\dfrac{1}{4.19}+...+\dfrac{1}{1900.2005}\) \(\&\) \(B=\dfrac{1}{2.1991}+\dfrac{1}{3.1992}+\dfrac{1}{4.1993}+...+\dfrac{1}{16.2005}\)
Làm giúp mk zới thứ 6 thi zùi
chứng minh: \(\dfrac{1}{65}\)<\(\dfrac{1}{5^3}+\dfrac{1}{6^3}+....+\dfrac{1}{2023^3}< \dfrac{1}{40}\)
Tính các số hữu tỉ sau:
A=\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).....\left(1-\dfrac{1}{n+1}\right),n\)là số tự nhiên
B=\(\dfrac{1}{2000.1999}-\dfrac{1}{1999.1998}-\dfrac{1}{1998.1997}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
Nhờ các các bn giúp mik!khó quá mik lm hk ra ai giỏi toán giúp mik với!
b,
\(B=\frac{1}{2000.1999}-\frac{1}{1999.1998}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(1-\frac{1}{1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\frac{1998}{1999}\)
\(\Rightarrow B=\frac{1}{1999}-\frac{1}{2000}-\frac{1998}{1999}\)
\(\Rightarrow B=\left(\frac{1}{1999}-\frac{1998}{1999}\right)-\frac{1}{2000}\)
\(\Rightarrow B=\frac{-1997}{1999}-\frac{1}{2000}\)
A= (1-\(\dfrac{1}{2}\) ) . ( 1- \(\dfrac{1}{3}\) ) . ( 1 - \(\dfrac{1}{4}\)) ...........( 1- \(\dfrac{1}{n+1}\))
= \(\dfrac{1}{2}\). \(\dfrac{2}{3}\) . \(\dfrac{3}{4}\) ......\(\dfrac{1}{n+1}\)
= 1
Mik nghĩ chắc là thê này đó Khánh Linh
a)
Có:
\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{n+1}\right)\)
\(\Leftrightarrow A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{n}{n+1}\)
Rút gọn biểu thức A, ta được:
\(A=\dfrac{1}{n-1}\)
Chúc bạn học tốt!
tính:
a) \(\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{4}-1\right)\cdot...\left(\dfrac{1}{1000}-1\right)\)
b) \(\dfrac{3}{^22}\cdot\dfrac{8}{^23}\cdot\dfrac{15}{^24}\cdot...\dfrac{99}{^210}\)
giúp mik với nhék! mik đang cần gấp.