Chứng minh:
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}< 2\)
Cho S = \(\dfrac{6}{15}+\dfrac{6}{16}+\dfrac{6}{17}+\dfrac{6}{18}+\dfrac{6}{19}\) Chứng minh rằng 1<S<2
Ta có \(\dfrac{6}{15}>\dfrac{6}{16}>...>\dfrac{6}{19}\) nên \(S< \dfrac{6}{15}.5=2\).
Lại có \(S>\dfrac{6}{19}.5>1\) nên \(1< S< 2\)
\(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{2}{18}+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}}\)
\(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)
Biến đổi tử số
\(19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}\)
= 1 + \(\left(1+\dfrac{18}{2}\right)+\left(1+\dfrac{17}{3}\right)+\left(1+\dfrac{16}{4}\right)+...+\left(1+\dfrac{1}{19}\right)\)
= \(\dfrac{20}{20}+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{1}{19}\)
= 20 x \(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)\)
Vậy \(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)
= \(\dfrac{20\times\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}=20\)
Vậy A = 20
Chứng minh rằng: \(S=\dfrac{1}{19}+\dfrac{1}{19^2}+\dfrac{1}{19^3}+...+\dfrac{1}{19^{10}}< \dfrac{1}{18}\)
`S=1/19+1/19^2+1/19^3+........+1/19^20`
`=>19S=1+1/19+1/19^2+.....+1/19^19`
`=>19S-S=18S=1-1/19^20<1`
`=>S<1/18(đpcm)`
Giải:
S=\(\dfrac{1}{19}+\dfrac{1}{19^2}+\dfrac{1}{19^3}+...+\dfrac{1}{19^{10}}\)
19S=\(1+\dfrac{1}{19}+\dfrac{1}{19^2}+...+\dfrac{1}{19^9}\)
19S-S=\(\left(1+\dfrac{1}{19}+\dfrac{1}{19^2}+...+\dfrac{1}{19^9}\right)-\left(\dfrac{1}{19}+\dfrac{1}{19^2}+\dfrac{1}{19^3}+...+\dfrac{1}{19^{10}}\right)\)
18S=1-\(\dfrac{1}{19^{10}}\)
S=(1-\(\dfrac{1}{19^{10}}\) ):18
S=\(1:18-\dfrac{1}{19^{10}}:18\)
S=\(\dfrac{1}{18}-\dfrac{1}{19^{10}.18}\)
⇒S<\(\dfrac{1}{18}\) (đpcm)
Chúc bạn học tốt!
S = 119+1192+1193+........+11920S=119+1192+1193+........+11920
⇒ 19S=1+119+1192+.....+11919⇒19S=1+119+1192+.....+11919
⇒ 19S−S=18S=1−11920<1⇒19S-S=18S=1-11920<1
⇒ S<118(đpcm)
1)Cho B=\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\).Chứng minh B>1
2)Tính nhanh:M=\(\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\)
2,
\(M=\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\) =\(\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)
\(=\dfrac{3}{4}\)
Tính:
\(\dfrac{\dfrac{1}{19}+\dfrac{2}{18}+\dfrac{3}{17}+.....\dfrac{18}{2}+\dfrac{19}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+....+\dfrac{1}{19}+\dfrac{1}{20}}\)
Ta có: \(\dfrac{1}{19}+\dfrac{2}{18}+...+\dfrac{19}{1}=\left(\dfrac{1}{19}+1\right)+\left(\dfrac{2}{18}+1\right)+...+1\)
\(=\dfrac{20}{19}+\dfrac{20}{18}+...+\dfrac{20}{2}+\dfrac{20}{20}=20\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}\right)\)
Thế lại bài toán ta được
\(\dfrac{\dfrac{1}{19}+\dfrac{2}{18}+...+\dfrac{19}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}=\dfrac{20\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}=20\)
Ta có
\(\dfrac{1}{19}+\dfrac{2}{18}+\dfrac{3}{17}+...+\dfrac{19}{1}\\ =\dfrac{1}{19}+1+\dfrac{2}{18}+1+\dfrac{3}{17}+1+...+\dfrac{19}{1}+1-19\\ =\dfrac{20}{19}+\dfrac{20}{18}+\dfrac{20}{17}+...+\dfrac{20}{1}-19\\ =\dfrac{20}{19}+\dfrac{20}{18}+...+\dfrac{20}{2}+20-19\\ =\dfrac{20}{19}+\dfrac{20}{18}+\dfrac{20}{17}+...+\dfrac{20}{2}+1+19-19\\ =\dfrac{20}{20}+\dfrac{20}{19}+\dfrac{20}{18}+...+\dfrac{20}{2}\\ =20\cdot\left(\dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{2}\right)\)
Thế vào ta có:
\(\dfrac{\dfrac{1}{19}+\dfrac{2}{18}+\dfrac{3}{17}+...+\dfrac{19}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\\ =\dfrac{20\cdot\left(\dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{2}\right)}{\dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{2}}\\ =20\)
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: \(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\)
\(=\dfrac{3-2-1}{6}\)
\(=0\)
\(\rightarrow C=0\)
Chứng minh:
C = \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
1/5<2;1/6<2;...;1/17<2
suy ra 1/5+1/6+1/7+...+1/17<2
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\)
\(=\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{10}\right)+\left(\dfrac{1}{11}+...+\dfrac{1}{17}\right)< \dfrac{1}{5}.6+\dfrac{1}{11}.7=\dfrac{6}{5}+\dfrac{7}{11}\)
\(=1\dfrac{46}{55}< 2\)
\(\Rightarrow dpcm\)
Bài 1.( 2 điểm)Tính bằng cách hợp lí:
a) \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
b) \(\left(\dfrac{18}{23}+\dfrac{7}{12}\right)+\left(\dfrac{-13}{19}-\dfrac{3}{4}\right)+\left(\dfrac{-6}{19}+\dfrac{5}{23}\right)\)
c) \(\dfrac{4}{3}+\dfrac{-5}{6}+\dfrac{-1}{4}\)
d) \(\dfrac{5}{6}-\dfrac{7}{5}+\dfrac{17}{30}\)
1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)
\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)
\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)
=1
chứng minh rằng
A= \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
B=\(\dfrac{5}{11}+\dfrac{5}{12}+\dfrac{5}{13}+\dfrac{5}{14},1< B< 2\)