Cho S = \(\dfrac{6}{15}+\dfrac{6}{16}+\dfrac{6}{17}+\dfrac{6}{18}+\dfrac{6}{19}\) Chứng minh rằng 1<S<2
Chứng minh:
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}< 2\)
Tính hợp lý:
\(a.\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}+\dfrac{-8}{13}\)
\(b.\dfrac{-3}{8}\cdot\dfrac{1}{6}+\dfrac{3}{-8}\cdot\dfrac{5}{6}+\dfrac{-10}{16}\)
\(c.\dfrac{-4}{11}\cdot\dfrac{5}{15}\cdot\dfrac{11}{-4}\)
a: \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}-\dfrac{8}{13}\right)+\left(\dfrac{-18}{35}-\dfrac{17}{35}\right)\)
=1-1-1
=-1
b: \(=\dfrac{-3}{8}\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}=\dfrac{-3}{8}-\dfrac{5}{8}=-1\)
c: \(=\dfrac{4}{4}\cdot\dfrac{5}{15}\cdot\dfrac{11}{11}=\dfrac{1}{3}\)
a) \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(-\dfrac{5}{13}+-\dfrac{8}{13}\right)+\left(-\dfrac{18}{35}+-\dfrac{17}{35}\right)=1+-1+-1=-1\)
b) \(=-\dfrac{3}{8}\cdot\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}-\dfrac{10}{16}=-1\)
c) \(=\left(-\dfrac{4}{11}\cdot-\dfrac{11}{4}\right)\cdot\dfrac{5}{15}=1\cdot\dfrac{1}{3}=\dfrac{1}{3}\)
a)\(=\left(-\dfrac{5}{13}+\dfrac{-8}{13}\right)+\left(-\dfrac{18}{35}-\dfrac{17}{35}\right)+\left(\dfrac{3}{14}+\dfrac{14}{17}\right)=-1-1+1=-1\)
b)\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}.1-\dfrac{10}{16}=-\dfrac{6}{16}-\dfrac{10}{16}=-\dfrac{16}{16}=-1\)
c)\(\dfrac{-4.5.11}{11.5.3.-4}=\dfrac{1}{3}\)
Bài 3:
c) Chứng tỏ rằng \(\dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}+...+\dfrac{1}{43}+\dfrac{1}{44}>\dfrac{5}{6}\)
Giúp mik vs! Thanks nhiều nha!
tính
\(\dfrac{27}{15}\)+\(\dfrac{6}{8}\)
\(\dfrac{19}{24}\)+\(\dfrac{7}{18}\)
\(\dfrac{2}{9}\)-\(\dfrac{1}{6}\)
\(\dfrac{8}{15}\)-\(\dfrac{1}{3}\)
27/15 + 6/8 =51/20
19/24 + 7/18 = 85/72
2/9 - 1/6 = 1/18
8/15 -1/3 = 1/5
\(15\dfrac{5}{6}+\left(\dfrac{17}{18}-\dfrac{5}{12}\right):3\dfrac{1}{6}\)
tính
f\(\dfrac{-2}{17}+\dfrac{15}{23}+\dfrac{-15}{17}+\dfrac{4}{19}+\dfrac{8}{23}\)
g \(\dfrac{-1}{2}+\dfrac{3}{21}+\dfrac{-2}{6}+\dfrac{-5}{30}\)
f)\(-\dfrac{2}{17}+\dfrac{15}{23}+-\dfrac{15}{17}+\dfrac{4}{19}+\dfrac{8}{23}\)
\(=\left(-\dfrac{2}{17}+-\dfrac{15}{17}\right)+\left(\dfrac{15}{23}+\dfrac{8}{23}\right)+\dfrac{4}{19}\)
\(=-1+1+\dfrac{4}{19}\)
\(=0+\dfrac{4}{19}=\dfrac{4}{19}\)
g)\(-\dfrac{1}{2}+\dfrac{3}{21}+-\dfrac{2}{6}+-\dfrac{5}{30}\)
\(=-\dfrac{1}{2}+\dfrac{1}{7}+-\dfrac{1}{3}+-\dfrac{1}{6}\)
\(=\left(-\dfrac{1}{2}+-\dfrac{1}{3}+-\dfrac{1}{6}\right)+\dfrac{1}{7}\)
\(=-\dfrac{3+2+1}{6}+\dfrac{1}{7}\)
\(=\dfrac{1}{7}-1\)
\(=\dfrac{1}{7}-\dfrac{7}{7}=-\dfrac{6}{7}\)
`f)(-2)/17 + 15/23 + (-15)/17 + 4/19 + 8/23`
`= (-2/17+ -15/17)+(15/23+8/23)+4/19`
`= -1+1+4/19`
`= 0 +4/19`
`= 0`
`g)(-1)/2 + 3/21 + (-2)/6 + (-5)/30`
`= (-1)/2 + 1/7 + (-1)/3 + (-1)/6`
`= (-21)/42 + 6/42 + (-14)/42 + (-7)/42`
`=(-36)/42`
`=(-6)/7`
\(f,-\dfrac{2}{17}+\dfrac{15}{23}-\dfrac{15}{17}+\dfrac{4}{19}+\dfrac{8}{23}\)
\(=-\dfrac{2}{17}-\dfrac{15}{17}+\dfrac{15}{23}+\dfrac{8}{23}+\dfrac{4}{19}\)
\(=-1+1+\dfrac{4}{19}\)
\(=\dfrac{4}{19}\)
\(g,-\dfrac{1}{2}+\dfrac{3}{21}-\dfrac{2}{6}-\dfrac{5}{30}\)
\(=-\dfrac{5}{14}-\dfrac{1}{2}\)
\(=-\dfrac{5}{14}-\dfrac{7}{14}\)
\(=-\dfrac{6}{7}\)
tính nhanh
C=[\(^{\dfrac{17}{28}}\)+\(\dfrac{18}{29}\)-\(\dfrac{19}{30}\)-\(\dfrac{20}{11}\)].[\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\)]
Xét: \(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\)
\(=\dfrac{3-2-1}{6}\)
\(=0\)
\(\rightarrow C=0\)
Cho biểu thức S=\(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{63}\)
Chứng minh rằng 3<S<6
Để chứng minh 3<S<6, ta cần tính giá trị của biểu thức S và thấy xem nó có nằm trong khoảng (3, 6) hay không.
Đầu tiên, ta tính tổng S bằng cách đặt S bên cạnh tổng harmonic thứ 63, rồi trừ đi tổng harmonic thứ 62:
S = 1/1 + 1/2 + 1/3 + ... + 1/63 S - 1/2 = 1/2 + 1/3 + ... + 1/63
Lặp lại phương pháp trên đối với S - 1/2, ta có:
S - 1/2 - 1/3 = 1/3 + ... + 1/63
Cứ lặp lại phương pháp trên đến khi ta được:
S - 1/2 - 1/3 - ... - 1/62 = 1/63
Tổng quát lại, ta có:
S - 1/2 - 1/3 - ... - 1/62 - 1/63 = 0
Từ đây suy ra:
3/2 < 1/2 + 1/3 + ... + 1/62 + 1/63 < 1 + 1/2 + 1/3 + ... + 1/62 < 6
Vì vậy, ta có:
3 < S < 6
Vậy, ta đã chứng minh được rằng 3<S<6.
Bài 1:
a, Tính \(\dfrac{5.4^{15}.9^{9} - 4.3^{20}.8^{9}}{5.2^{9}.6^{19}-7.2^{29}.27^{6}}\)
b,Tìm x biết:
\(1\dfrac{1}{30}:(24\dfrac{1}{6}-24\dfrac{1}{5}) -\dfrac{1\dfrac{1}{2}-\dfrac{3}{4}}{4x-\dfrac{1}{2}}=(-1\dfrac{1}{15}):(8\dfrac{1}{5}-8\dfrac{1}{3})\)
Bài 2: Chứng minh số:
222...22200333...333(2001c/s 2; 2003 c/s3)