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30 tháng 4 2016

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}\)

\(S=1-\frac{1}{46}<1\)

=>chứng minh bị sai hoặc đề sai

30 tháng 4 2016

S=\(\frac{3}{1.4}+\frac{3}{4.7}+...........+\frac{3}{43.46}\)

=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...........+\frac{1}{43}-\frac{1}{46}\)

=\(1-\frac{1}{46}<1\)

\(\Rightarrow S<1\)

23 tháng 4 2016

Cho S=3/1x4+3/4x7+3/7x10+...+3/40x43+3/43x46. Hãy chứng tỏ S<1

ĐPM : S < 1

23 tháng 4 2016

S=3/1x4+3/4x7+3/7x10+...+3/40x43+3/43x46

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}\)

\(S=1-\frac{1}{46}\)

=>S<1

1 tháng 5 2015

= 1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

= 1 - 1/46 = 45/46 < 1

21 tháng 4 2016

S=3/1.4+3/4.7+3/7.10+.....+3/40.43+3/43.46

S= 1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

S= 1-1/46

=> S<1

24 tháng 4 2016

S=3.(1/1-1/4+1/4-1/7+.........+1/40-1/43+1/43-1/46)          

S=3.(1/1-1/46)

S=3.45/46

S=2/43/46

=> 2/43/46>1

=>S>1

6 tháng 6 2021

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}=1-\frac{1}{46}=\frac{45}{46}\)

6 tháng 6 2021

Trả lời:

\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}+\frac{3}{43\cdot46}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(=\frac{1}{1}-\frac{1}{46}\)

\(=\frac{46}{46}-\frac{1}{46}\)

\(=\frac{45}{46}\)

21 tháng 1 2020

Ta có:

S=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)

S=\(1-\frac{1}{n+3}\)

=>S<1

Vậy S<1

\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)

21 tháng 1 2020

Sory mình bấm bị lỗi

27 tháng 6 2017

\(A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{40.43}\)

\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.....+\dfrac{1}{40}-\dfrac{1}{43}\)

\(A=1-\dfrac{1}{43}\)

\(A< 1\left(đpcm\right)\)

27 tháng 6 2017

\(A=3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}\right)\)

\(=3\left(1-\dfrac{1}{43}\right)=\dfrac{126}{43}>1\)

... sai đâu không nhỉ??

=1-1/4+1/4-1/7+1/7-...+1/37-1/40

=1-1/40=39/40

21 tháng 5 2022

Mình cần gấp ạ. Mốt mik thi rồi