\(chob=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}.hãychứngtỏB>1\)
1) Tính:
a) \(\frac{\left(1+\frac{17}{1}\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right).....\left(1+\frac{17}{19}\right)}{\left(1+\frac{19}{1}\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right).....\left(1+\frac{19}{17}\right)}\)
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}\)
c) \(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}\)
e) \(\frac{\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2017}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}\)
2) CMR: \(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{800}}< \frac{1}{3}\)
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) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)
Tính:
a) \(\frac{\left(1+\frac{17}{1}\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right).....\left(1+\frac{17}{19}\right)}{\left(1+\frac{19}{1}\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right).....\left(1+\frac{19}{17}\right)}\)
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}\)
c) \(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}\)
e) \(\frac{\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2017}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}\)
2) CMR: \(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{800}}< \frac{1}{3}\)
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 \(\frac{1}{2^2}+\frac{1}{2^4}+........+\frac{1}{2^{800}}< \frac{1}{3}\)
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:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(=1+\frac{1}{2}-2.\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-2.\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-2.\frac{1}{2016}+\frac{1}{2017}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{2016}\right)\)
\(\left(-1\frac{1}{6}\right)\left(\frac{1-\frac{3}{5}+\frac{3}{11}-\frac{3}{13}}{\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}}\right):\left(\frac{4-\frac{4}{17}+\frac{4}{19}-\frac{4}{2013}}{5-\frac{5}{17}+\frac{5}{19}-\frac{5}{2013}}\right)\)
tính
\(\left(-1\frac{1}{6}\right)\left(\frac{1-\frac{3}{5}+\frac{3}{11}-\frac{3}{13}}{\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}}\right)\left(\frac{4-\frac{4}{17}+\frac{4}{19}-\frac{4}{2013}}{5-\frac{5}{7}+\frac{5}{19}-\frac{5}{2013}}\right)\)
\(=-\frac{7}{6}.\left(\frac{3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}\right)}{\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}}\right):\left(\frac{4.\left(1-\frac{1}{7}+\frac{1}{19}-\frac{1}{2013}\right)}{5.\left(1-\frac{1}{7}+\frac{1}{19}-\frac{1}{2013}\right)}\right)\)
\(=-\frac{7}{6}.3:\frac{4}{5}=-\frac{7}{2}.\frac{5}{4}=-\frac{35}{8}\)
\(\frac{\frac{4}{17}+\frac{4}{19}-\frac{4}{2111}}{\frac{5}{17}+\frac{5}{19}-\frac{5}{2111}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{\frac{-5}{123}+\frac{5}{19}-\frac{5}{371}+1}\)
\(\frac{\frac{4}{17}+\frac{4}{19}-\frac{4}{2111}}{\frac{5}{17}+\frac{5}{19}-\frac{5}{2111}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{-\frac{5}{123}+\frac{5}{19}-\frac{5}{371}+1}\)
\(=\frac{4.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}{5.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}+\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{5.\left(\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}\right)}=\frac{4}{5}+\frac{1}{5}=1\)
Cho tam giác ABC có đường cao AD .Gọi E là trung điểm của AB .F đối xứng vs D qua E c/m AB = DF
\([\frac{1\frac{11}{31}.4\frac{3}{7}-\left(15-6\frac{1}{3}.\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}.\left(-1\frac{14}{93}\right)].\frac{31}{50}\)
[\(\frac{-75}{59}\).\(\frac{-107}{93}\)]\(\frac{31}{50}\)=\(\frac{2675}{1829}\).\(\frac{31}{50}\)=\(\frac{107}{118}\)
\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{1}{6}\cdot\frac{20}{3}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-15+\frac{2}{3}}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\left(-\frac{150}{107}\right)\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}=\frac{50}{31}\cdot\frac{31}{50}=1\)
Tinh B=1\(\frac{6}{41}\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2006}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2006}}.\frac{124242423}{2373735}\)
\(\frac{5}{23}x\frac{17}{6}+\frac{5}{23}x\frac{9}{26}\)
\(=\frac{5}{23}x\left(\frac{17}{26}+\frac{9}{26}\right)\)
\(=\frac{5}{23}x1=\frac{5}{23}\)
\(a.A=[\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}]+\frac{1890}{2005}+115\)
b.B=\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\cdot\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(42-5\frac{1}{3}\right)}\cdot\left(-1\frac{19}{93}\right)\right]\cdot\frac{31}{50}\)
Tính
\(\frac{1\frac{11}{31}.4\frac{3}{7}-\left(15.6\frac{1}{3}.\frac{2}{19}\right)}{-4\frac{5}{6}+\frac{1}{6}.\left(12-5\frac{1}{3}\right)}.\left(-1\frac{14}{93}\right).\frac{31}{56}\)
Tính \(E=\left[\frac{1\frac{11}{31}.4\frac{3}{7}-\left(15-6\frac{1}{3}.\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}.\left(-1\frac{14}{93}\right)\right].\frac{31}{50}\)