a, Tính bằng cách hợp lí :
1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/110+1/132
b, Chứng tỏ rằng :
1/501+1/502+1/503+........+1/1000 <1
K
Khách
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Những câu hỏi liên quan
TM
1
S
1
19 tháng 6 2015
\(C=\frac{8}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{8}{90}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)
\(=\frac{8}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(=\frac{4}{45}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{4}{45}-\left(1-\frac{1}{9}\right)=\frac{4}{45}-\frac{8}{9}=\frac{4}{45}-\frac{40}{45}=\frac{-36}{45}=\frac{-4}{5}\)
PP
3
TH
Thầy Hùng Olm
Manager
VIP
4 tháng 6 2023
\(\dfrac{1}{20}=\dfrac{1}{4x5}=\dfrac{1}{4}-\dfrac{1}{5}\)
Tương tự các phân số khác
S= \(\dfrac{1}{4}-\dfrac{1}{12}=\dfrac{1}{6}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
4 tháng 6 2023
\(\dfrac{1}{20}+\dfrac{1}{30}\)+ \(\dfrac{1}{42}\)+\(\dfrac{1}{56}\)+\(\dfrac{1}{72}\)+\(\dfrac{1}{90}\)+\(\dfrac{1}{110}\)+\(\dfrac{1}{132}\)
= \(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+\(\dfrac{1}{6\times7}\)+\(\dfrac{1}{7\times8}\)+\(\dfrac{1}{8\times9}\)+\(\dfrac{1}{9\times10}\)+\(\dfrac{1}{10\times11}\)+\(\dfrac{1}{11\times12}\)
= \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)-\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{12}\)
= \(\dfrac{1}{4}\) - \(\dfrac{1}{12}\)
= \(\dfrac{3}{12}\) - \(\dfrac{1}{12}\)
= \(\dfrac{2}{12}\)
=\(\dfrac{1}{6}\)
NT
2
D
29 tháng 5 2022
`=1/[4xx5]+1/[5xx6]+1/[6xx7]+...+1/[11xx12]`
`=1/4-1/5+1/5-1/6+1/6-1/7+...+1/11-1/12`
`=1/4-1/12=3/12-1/12=2/12=1/6`
29 tháng 5 2022
\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\\ =\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+\dfrac{1}{8\times9}+\dfrac{1}{9\times10}+\dfrac{1}{10\times11}+\dfrac{1}{11\times12}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\\ =\dfrac{1}{4}-\dfrac{1}{12}\\ =\dfrac{3}{12}-\dfrac{1}{12}=\dfrac{2}{12}=\dfrac{1}{6}\)
NH
1
2 tháng 5 2023
Ta viết lại biểu thức A như sau:
\(A=-\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{11.12}\right)\)
\(A=-\left(\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{12-11}{11.12}\right)\)
\(A=-\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{12}\right)\)
\(A=-\left(\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(A=-\dfrac{1}{6}\)
BH
1
20 tháng 7 2016
Đặt tổng trên là A ta có :
A= 1/ 20 + 1/ 30 + 1/ 42 + 1/ 56 + 1/ 72 + 1/90 + 1/110 + 1 / 123 + 1/ 156
= 1 / 4 x5 + 1/ 5 x 6 + 1/6x 7 + 1/ 7x8 + 1/8x9 + 1/9x10+ 1/ 10x11+ 1 /11x12 +1/12 x 13
= 1/4- 1/5 + 1/ 5 - 1/6 + 1/ 6 - 1/7 + 1/ 7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10+ 1/10 - 1/11 + 1/11 - 1/12+ 1/ 12 - 1/13
= 1 /4 - 1 /13
= 9 /52
BN
6 tháng 4 2017
Ta có: \(\dfrac{1}{501}< \dfrac{1}{500}\)
\(\dfrac{1}{502}< \dfrac{1}{500}\)
\(\dfrac{1}{503}< \dfrac{1}{500}\)
..................
\(\dfrac{1}{1000}< \dfrac{1}{500}\)
\(\Rightarrow\dfrac{1}{501}+\dfrac{1}{502}+\dfrac{1}{503}+...+\dfrac{1}{1000}< \dfrac{1}{500}+\dfrac{1}{500}+\dfrac{1}{500}+...+\dfrac{1}{500}\)
\(\Rightarrow\dfrac{1}{501}+\dfrac{1}{502}+\dfrac{1}{503}+...+\dfrac{1}{1000}< \dfrac{500}{500}=1\)
Vậy \(\dfrac{1}{501}+\dfrac{1}{502}+\dfrac{1}{503}+...+\dfrac{1}{1000}< 1\)
6 tháng 4 2017
Đặt A = \(\dfrac{1}{501}+\dfrac{1}{502}+\dfrac{1}{503}+...+\dfrac{1}{1000}\)
Ta thấy A có 500 phân số.
Ta có: \(\dfrac{1}{501}< \dfrac{1}{500}\\ \dfrac{1}{502}< \dfrac{1}{500}\)
....................
\(\dfrac{1}{1000}< \dfrac{1}{500}\)
\(\Rightarrow\) A< \(\dfrac{1}{500}+\dfrac{1}{500}+...+\dfrac{1}{500}\)( có 500 phân số \(\dfrac{1}{500}\))
\(\Rightarrow A< 500.\dfrac{1}{500}\\ \Rightarrow A< \dfrac{500}{500}\\ \Rightarrow A< 1\)
Chắc là bạn hiểu chứ ?
KM
4
8 tháng 7 2017
\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{132}=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{11\cdot12}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{3}-\frac{1}{12}=\frac{4}{12}-\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)
Chú ý: \(\cdot=\times\)
DP
8 tháng 7 2017
Đặt \(A=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\)
\(A=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)
\(A=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{12}=\frac{1}{4}\)
KL
3
TA
23 tháng 7 2019
\(=\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{12.13}\)
áp dụng \(\frac{1}{a.b}=\frac{1}{a}-\frac{1}{b}\)làm sẽ có các số nghịch đảo và được kết quả là 1/4 - 1/13
23 tháng 7 2019
A = 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/110 + 1/132 + 1/156
A = 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 + 1/10.11 + 1/11.12 + 1/12.13
A = 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10 + 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13
A = 1/4 - 1/13
A = 9/52
a)Đặt \(A=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)
\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\)
\(A=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)
\(A=\dfrac{1}{3}-\dfrac{1}{12}\)
\(A=\dfrac{1}{4}\)
b)Đặt \(B=\dfrac{1}{501}+\dfrac{1}{502}+...+\dfrac{1}{1000}\)(có 500 số hạng)
\(B< \dfrac{1}{500}+\dfrac{1}{500}+...+\dfrac{1}{500}\)(có 500 số hạng)
\(B< 500\cdot\dfrac{1}{500}=1\)
\(\Rightarrow B< 1\left(đpcm\right)\)
bạn có thể chỉ rõ 500 số hạng ở đâu ko