Chứng minh rằng :
\(\dfrac{1}{3}+\dfrac{1}{31}+\dfrac{1}{35}+\dfrac{1}{37}+\dfrac{1}{47}+\dfrac{1}{53}+\dfrac{1}{61}< \dfrac{1}{2}\)
Thực hiện phép tính một cách hợp lý :
a) \(\dfrac{-12}{7}\) . \(\dfrac{4}{35}\) + \(\dfrac{12}{7}\) . \(\dfrac{-31}{35}\) - \(\dfrac{2}{7}\)
b) 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... + 97 + 98 - 99 - 100.
c) A = 157 . ( - 37 ) - ( 41 . 53 - 37 . 157 ) + 51 . 53
d) B = \(\left(\dfrac{1}{11}+\dfrac{1}{21}+\dfrac{1}{31}+\dfrac{1}{41}+\dfrac{1}{51}\right)\) \(\left(\dfrac{-41}{123}+\dfrac{31}{-186}-\dfrac{-51}{102}\right)\)
a: \(=\dfrac{-12}{7}\left(\dfrac{4}{35}+\dfrac{31}{35}\right)-\dfrac{2}{7}=\dfrac{-12}{7}-\dfrac{2}{7}=-2\)
b: =(-4)+(-4)+...+(-4)
=-4*25=-100
c: \(=157\cdot\left(-37\right)-41\cdot53+37\cdot157+51\cdot53\)
=10*53
=530
CHỨNG TỎ RẰNG :
\(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{33}\)+\(\dfrac{1}{35}\)+.....+\(\dfrac{1}{148}\)+\(\dfrac{1}{149}\)\(\dfrac{1}{150}\)<\(\dfrac{13}{6}\)
BT2: Tính nhanh:
1) \(\dfrac{15}{37}\left(\dfrac{38}{41}-\dfrac{74}{45}\right)-\dfrac{38}{41}\left(\dfrac{15}{37}+\dfrac{82}{76}\right)\)
2)\(\dfrac{47}{53}\left(\dfrac{17}{3}-\dfrac{53}{47}\right)+\dfrac{17}{3}\left(\dfrac{6}{17}-\dfrac{47}{53}\right)\)
1: \(=\dfrac{15}{37}\cdot\dfrac{38}{41}-\dfrac{15}{37}\cdot\dfrac{74}{45}-\dfrac{38}{41}\cdot\dfrac{15}{37}-\dfrac{38}{41}\cdot\dfrac{82}{76}\)
\(=\dfrac{-2}{3}-1=-\dfrac{5}{3}\)
2: \(=\dfrac{47}{53}\cdot\dfrac{17}{3}-\dfrac{47}{53}\cdot\dfrac{53}{47}+\dfrac{17}{3}\cdot\dfrac{6}{17}-\dfrac{17}{3}\cdot\dfrac{47}{53}\)
\(=-1+2=1\)
\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)
\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)
\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)
\(=\dfrac{-6}{5}.4:\dfrac{4}{5}\)
\(=\dfrac{-6.4.5}{5.4}=-6\)
\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)
\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)
\(=\dfrac{-6}{5}.\dfrac{4}{3}:\dfrac{4}{5}\)
\(=\dfrac{-6.4.5}{5.3.4}=\dfrac{-6}{3}=-2\)
Vậy...
\(l,\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}\)
\(m,\left(3-2\dfrac{1}{3}+\dfrac{1}{4}\right):\left(4-5\dfrac{1}{6}+2\dfrac{1}{4}\right)\)
\(n,F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
\(p,F=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)
p: \(F=\dfrac{1}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+\dfrac{3}{9\cdot12}+...+\dfrac{3}{30\cdot33}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)
n: \(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)
m: \(=\left(3-\dfrac{7}{3}+\dfrac{1}{4}\right):\left(4-\dfrac{31}{6}+\dfrac{9}{4}\right)\)
\(=\dfrac{36-28+3}{12}:\dfrac{48-62+27}{12}\)
\(=\dfrac{11}{13}\)
Chứng minh :
A =\(\dfrac{1}{2}+\dfrac{1}{33}+\dfrac{1}{34}+\dfrac{1}{35}+\dfrac{1}{51}+\dfrac{1}{53}+\dfrac{1}{55}+\dfrac{1}{57}+\dfrac{1}{59}\)<\(\dfrac{7}{10}\)
Lời giải:
\(A=\frac{1}{2}+\frac{1}{33}+\frac{1}{34}+\frac{1}{35}+\frac{1}{51}+\frac{1}{53}+\frac{1}{55}+\frac{1}{57}+\frac{1}{59}\)
Ta có:
\(\frac{1}{33}+\frac{1}{34}+\frac{1}{35}< \frac{1}{30}+\frac{1}{30}+\frac{1}{30}=\frac{3}{30}=\frac{1}{10}\)
\(\frac{1}{51}+\frac{1}{53}+\frac{1}{55}+\frac{1}{57}+\frac{1}{59}< \frac{1}{50}+\frac{1}{50}+\frac{1}{50}+\frac{1}{50}+\frac{1}{50}=\frac{5}{50}=\frac{1}{10}\)
Cộng theo vế:
\(\frac{1}{33}+\frac{1}{34}+\frac{1}{35}+\frac{1}{51}+\frac{1}{53}+\frac{1}{55}+\frac{1}{57}+\frac{1}{59}< \frac{2}{10}=\frac{1}{5}\)
Suy ra \(A< \frac{1}{2}+\frac{1}{5}=\frac{7}{10}\)
Ta có đpcm.
Chứng minh rằng:
\(\dfrac{1}{5}-\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}< \dfrac{1}{2}\)
Sai đề. Sửa đề :v
Cmr: \(\dfrac{1}{5}+\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}< \dfrac{1}{2}\)
Giải:
Đặt \(A=\dfrac{1}{5}+\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}\)
Ta có:
\(A=\dfrac{1}{5}+\left(\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}\right)+\left(\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}\right)\)
\(A< \dfrac{1}{5}\left(\dfrac{1}{14.3}\right)+\left(\dfrac{1}{61.3}\right)\)
\(A< \dfrac{1}{5}+\dfrac{3}{14}+\dfrac{3}{61}\)
\(A< \dfrac{1}{5}+\dfrac{3}{12}+\dfrac{1}{20}\)
\(A< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)
\(\Rightarrow A< \dfrac{1}{2}\)
Vậy \(\dfrac{1}{5}+\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}< \dfrac{1}{2}\) \((đpcm)\)
\(a,\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}+\dfrac{-0,25.\dfrac{-2}{3}-75\%:\left(\dfrac{-1}{2}+\dfrac{2}{3}\right)}{\left|-1\dfrac{1}{2}\right|.\left(\dfrac{-2}{3}-0,75:\dfrac{3}{-2}\right)}\)
Chứng minh rằng:
S=\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{2}\)
Giải:
Ta có:
\(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)
\(=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\) \(\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)
Nhận xét:
\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{4}\)
\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)
\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)
Vậy \(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\) \(< \dfrac{1}{2}\) (Đpcm)