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DV
3
29 tháng 6 2015
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{9^2}-\frac{1}{10^2}=1-\frac{1}{10^2}
DN
3 tháng 3 2022
a) \(2^5+8\left[\left(-2\right)^3:\frac{1}{2}\right]^0-\left(\frac{1}{2}\right)^3\times2+\left(-2\right)^3\)
\(=32+8\times1-\frac{1}{8}\times2+\left(-8\right)\)
\(=32+8-\frac{1}{4}+\left(-8\right)\)
\(=40-\frac{1}{4}+\left(-8\right)\)
\(=39\frac{3}{4}+\left(-8\right)\)
\(=31\frac{3}{4}\)
b vaf c mai minhf lamf, ht
EC
1 tháng 9 2016
a)|-10|:(-2):(-5)+(-3)2
=1+9
=10
b)1+(-2)+3+(-4)+5+(-6)+...+21+(-22)
=[1+(-2)]+[3+(-4)]+[5+(-6)]+...+[21+(-22]
=(-1)+(-1)+(-1)+...+(-1)
Mà từ 1 đến 22 có:(22-1):1+1:2=11(cặp)
Suy ra:1+(-2)+3+(-4)+5+(-6)+...+21+(-22)=(-11)
EC
1 tháng 9 2016
c)\(\frac{3}{4}.\frac{5}{9}+\frac{3}{4}.\frac{4}{9}\)
\(=\frac{3}{4}.\left(\frac{5}{9}+\frac{4}{9}\right)\)
\(=\frac{3}{4}\)
d)\(-\frac{4}{17}+\frac{5}{19}+-\frac{13}{17}+\frac{14}{19}+\frac{3}{115}\)
\(=\left[\left(-\frac{4}{17}\right)+\left(-\frac{13}{17}\right)\right]+\left(\frac{5}{19}+\frac{4}{19}\right)+\frac{3}{115}\)
\(=\left(-\frac{27}{17}\right)+1+\frac{3}{115}\)
\(=-\frac{1099}{1955}\)
e)\(\left(\frac{3}{4}+-\frac{7}{2}\right).\left(\frac{10}{11}+\frac{2}{22}\right)\)
\(=\left(\frac{3}{4}-\frac{14}{4}\right).\left(\frac{20}{22}+\frac{2}{22}\right)\)
\(=\left(-\frac{11}{4}\right).\left(\frac{22}{22}\right)\)
\(=-\frac{11}{4}\)
14 tháng 9 2016
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=\frac{-2}{3}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)
15 tháng 9 2016
Làm tiếp:
\(=\left(1+\frac{1}{2}+.....+\frac{1}{2017}\right)-\left(1+\frac{1}{2}+....+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+.........+\frac{1}{2017}\)
\(\Rightarrow\frac{\frac{1}{1009}+....+\frac{1}{2017}}{1-\frac{1}{2}+.....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}=1\)
Bài 2:
Đặt \(A=\frac{1}{2^2}+.......+\frac{1}{2^{800}}\)
\(4A=1+\frac{1}{2^2}+.....+\frac{1}{2^{798}}\)
\(\Rightarrow4A-A=1-\frac{1}{2^{800}}\)
\(\Rightarrow3A=1-\frac{1}{2^{800}}< 1\Rightarrow A< \frac{1}{3}\)
Vậy \(\frac{1}{2^2}+\frac{1}{2^4}+........+\frac{1}{2^{800}}< \frac{1}{3}\)
15 tháng 9 2016
Bài 1:Tính
a, Xét biểu thức \(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).........\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)..........\left(1+\frac{n+2}{n}\right)}\) với\(n\in N\)
Ta có:\(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).......\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)......\left(1+\frac{n+2}{n}\right)}\)
\(=\frac{\frac{n+1}{1}.\frac{n+2}{2}........\frac{2n+2}{n+2}}{\frac{n+3}{1}.\frac{n+4}{2}.........\frac{2n+2}{n}}\)
\(=\frac{\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right)}{1.2.3.........\left(n+2\right)}}{\frac{\left(n+3\right)\left(n+4\right)........\left(2n+2\right)}{1.2.3.........n}}\)
\(=\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right).1.2.3.......n}{\left(n+3\right)\left(n+4\right)........\left(2n+2\right).1.2.3......\left(n+2\right)}\)
\(=\frac{\left(n+1\right)\left(n+2\right)}{\left(n+1\right)\left(n+2\right)}=1\)
Áp dụng vào bài toán ta có đáp số là:1
b, \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=-\frac{2}{3}\)
c,\(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}{\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}=\frac{\frac{1}{3}}{\frac{1}{4}}=12\)
d,\(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}=\frac{2}{13}\)
e,Xét mẫu số ta có:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(=1+\frac{1}{2}-2.\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-2.\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-2.\frac{1}{2016}+\frac{1}{2017}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{2016}\right)\)
11 tháng 7 2016
c.\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(\frac{\frac{25}{108}.\frac{5751}{25}+\frac{187}{4}}{\frac{100}{21}:-\frac{41}{21}}\)
\(\frac{\frac{213}{4}+\frac{187}{4}}{-\frac{100}{41}}\)
\(\frac{100}{-\frac{100}{41}}=-41\)
11 tháng 7 2016
a. \(\frac{4}{9}:-\frac{1}{7}+6\frac{5}{9}:-\frac{1}{7}\)
\(\left(\frac{4}{9}+6\frac{5}{9}\right):-\frac{1}{7}\)
\(7:-\frac{1}{7}=-49\)