cho S= \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}+\dfrac{3}{43.46}\)
Hãy chứng tỏ rằng S<1
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}+\dfrac{3}{43.46}\\ S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\\ S=1-\dfrac{1}{46}< 1\)
Vậy S < 1 (đpcm)
Cho S=\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+......+\dfrac{3}{40.43}+\dfrac{3}{43.46}\).Hãy chứng tỏ S<1
\(S=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{43\cdot46}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{43}-\dfrac{1}{46}\)
\(S=1-\dfrac{1}{46}< 1\)
S= \(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{40\cdot43}+\dfrac{3}{43\cdot46}\)
S= \(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{42}-\dfrac{1}{46}\)
S= \(1-\dfrac{1}{46}\)
S= \(\dfrac{45}{46}\)
Mà \(\dfrac{45}{46}< 1\)
\(\Rightarrow S< 1\)
Vậy S < 1
Cho S=\(\dfrac{3}{1.4}\)+\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{43.46}\)
cmr S<1
Ta có:
\(S=\dfrac{3}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{43}-\dfrac{1}{46}\right)\)
\(S=1.\left(\dfrac{1}{1}-\dfrac{1}{46}\right)\)
\(S=1.\dfrac{45}{46}=\dfrac{45}{46}\)
Vì \(\dfrac{45}{46}< \dfrac{46}{46}\) nên \(\dfrac{45}{46}< 1\).
Vậy S < 1.
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{43.46}\)
\(S=\dfrac{3}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{43.46}\right)\)
Ta thấy:
\(\dfrac{3}{1.4}=1-\dfrac{1}{4};\dfrac{3}{4.7}=\dfrac{1}{4}-\dfrac{1}{7};\dfrac{3}{7.10}=\dfrac{1}{7}-\dfrac{1}{10};\)
\(...;\dfrac{3}{43.46}=\dfrac{1}{43}-\dfrac{1}{46}\)
\(\Rightarrow S=1\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{43}-\dfrac{1}{46}\right)\)
\(\Rightarrow S=1\left(1-\dfrac{1}{46}\right)\)
\(\Rightarrow S=1.\dfrac{45}{46}=\dfrac{45}{46}\)
Ta có S=1-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{10}\)+...+\(\dfrac{1}{43}\)-\(\dfrac{1}{46}\)
=1-\(\dfrac{1}{46}\)<1
=>S<1
\(Cho\) \(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{n\left(n+1\right)}\)
Chưng minh rằng S<1
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...................+\dfrac{3}{n\left(n+1\right)}\)
\(\Rightarrow S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.............+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\Rightarrow S=1-\dfrac{1}{n+1}< 1\)
\(\Rightarrow S< 1\rightarrowđpcm\)
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{n.\left(n+1\right)}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...-\dfrac{1}{n+1}\)
\(S=1-\dfrac{1}{n+1}\)\(< 1\)
\(\Leftrightarrow S< 1\)
tik cho mik nhé
\(S=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{40\cdot43}+\dfrac{3}{43\cdot46}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\)
\(=1-\dfrac{1}{46}< 1\)
Vậy S < 1 ( ĐPCM )
1.Tính :
a ) \(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2018.2019}\)
b ) \(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{2017.2019}\)
c) \(S=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+....+\dfrac{1}{2018.2020}\)
d) \(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{2017.2018.2019}\)
2. Tính tổng:
a) \(S=1.2+2.3+3.4+...+2018.2019\)
b) \(S=3.5+5.7+7.9+...+2017.2019\)
c) \(S=2.4+4.6+6.8+...+2018.2020\)
d) \(S=1.2.3+2.3.4+3.4.5+...+2017.2018.2019\)
3.Tính
a ) \(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.....+\dfrac{1}{2017.2020}\)
b ) \(S=\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+\dfrac{1}{5.7.9}+....+\dfrac{1}{2017.2019.2021}\)
Ai có công thức không cho mình xin với ????
Bài 1a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2018.2019}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)
b) \(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2017.2019}\)
\(2S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2017.2019}\)
\(2S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2019}\)
\(2S=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)
\(S=\dfrac{1009}{2019}\)
Còn lại bạn làm tương tự hết nhé .
Cho S= \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{n.\left(n+3\right)}\)(với n \(\in\) N*). Chứng tỏ rằng S<1.
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{n\left(n+3\right)}\)
\(\Rightarrow S=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{\left(n+3\right)-n}{n\left(n+3\right)}\)
\(\Rightarrow S=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{n+3}{n\left(n+3\right)}-\dfrac{n}{n\left(n+3\right)}\)
\(\Rightarrow S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{n}-\dfrac{1}{n+3}\)
\(\Rightarrow S=1-\dfrac{1}{n+3}< 1\Rightarrow S< 1\)
Vậy S < 1
Tính nhanh
a, S= \(\dfrac{1}{3}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{63}\) + \(\dfrac{1}{99}\) + \(\dfrac{1}{143}\)
b, A = \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\)
c, H =\(\dfrac{4047991-2010x2009}{4050000-2011x2009}\)
d, T = \(\dfrac{2009x20010+2000}{2011x2010-2020}\)
e, P = \(\dfrac{7589-80,5x69,3}{7485,05-79x69,3}\)
f, B = 5,1 x 42,2 + 1,7 x 448 x 3 - 0,15 x 700
Giúp mình với
a=78/35
b=22/12
c=1/1
d=40202090/4040090
e=1,24025667172...
f=871,82
ko biết đúng ko [0_0'] hihi
tính tổng: A= \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\) B= \(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{3.7}+...+\dfrac{5}{99.101}\)
C= \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\) D= \(\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\) E= \(\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)
=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)
=2.(1-1/101)
=2.(101/101-1/101)
=2.100/101
200/101
B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)
=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)
=2.(1/1+1/101)
=2.(101/101+1/101)
=2.102/101
=204/101
C=1/2+1/3+1/3+1/4+....+1/99+1/100
=1/2+1/100
=50/100+1/100
=51/100
So sánh A và B với \(\dfrac{1}{2}\) biết :
\(A=\dfrac{1}{1.2^2}+\dfrac{1}{2.3^2}+\dfrac{1}{3.4^2}+........+\dfrac{1}{49.50^2}\) và
\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.......+\dfrac{1}{50^2}\)
Ta thấy:
\(1\cdot2^2=2^2;2\cdot3^2>3^2;3\cdot4^2>4^2;...;49\cdot50^2>50^2\)
\(\Rightarrow\dfrac{1}{1.2^2}=\dfrac{1}{2^2};\dfrac{1}{2\cdot3^2}< \dfrac{1}{3^2};\dfrac{1}{3\cdot4^2}< \dfrac{1}{4^2};...;\dfrac{1}{49\cdot50^2}< \dfrac{1}{50^2}\)
\(\Rightarrow\dfrac{1}{1\cdot2^2}+\dfrac{1}{2\cdot3^2}+\dfrac{1}{3\cdot4^2}+...+\dfrac{1}{49\cdot50^2}< \dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)
hay A<B
Vậy A<B
