1 phần 3'2+1 phần 4'5+ 1 phần 5'2+...+1phần100'2<1phần 2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
10 tháng 5 2016
ta có : 2/3'2 < 2/2.3 ; 2/4'2<2/3.4 ... ;2/100'2<2/99.100
nen 2/3'2 +2/4'2+...+2/100'2<2/2.3+2/3.4+...+2/99.100 (1)
ta có 2/2.3+2/3.49+...+2/99.100
=2.(1/2-1/3+1/3-1/4+...+1/99-1/100)
=2.(1/2-1/100)
=2.(50/100-1/100)
=2.49/100
ma 1>49/100
nen 1>1/2-1/3+...+1/99-1/100 (2)
tu(1) va (2) suy ra 2/3'2+...+2/100'2 >1
25 tháng 8 2023
\(A=1^2+3^2+5^2+...+99^2\)
=>\(A=\left(1^2+2^2+...+99^2+100^2\right)-\left(2^2+4^2+...+100^2\right)\)
\(=\left(1^2+2^2+...+100^2\right)-4\left(1^2+2^2+...+50^2\right)\)
\(=\dfrac{100\cdot\left(100+1\right)\left(100\cdot2+1\right)}{6}-4\cdot\dfrac{50\cdot\left(50+1\right)\left(50\cdot2+1\right)}{6}\)
\(=166650\)
2 tháng 6 2018
a) \(\left(4\frac{1}{2}-2x\right)\cdot3\frac{2}{3}=\frac{11}{5}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{5}\cdot\frac{3}{11}\)
\(2x=\frac{45-6}{10}\)
\(2x=\frac{39}{10}\)
\(x=\frac{39}{10\cdot2}=\frac{39}{20}\)
b) \(\frac{3}{4}\cdot x+\frac{4}{7}\cdot x=-\frac{15}{8}\)
\(x\cdot\left(\frac{21+16}{28}\right)=-\frac{15}{8}\)
\(x=-\frac{15}{8}\cdot\frac{28}{37}\)
\(x=-\frac{105}{74}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
15 tháng 8 2023
A = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+...+ \(\dfrac{1}{812}\) + \(\dfrac{1}{870}\)
A = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{28\times29}\)+ \(\dfrac{1}{29\times30}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+\(\dfrac{1}{28}\)-\(\dfrac{1}{29}\)+ \(\dfrac{1}{29}\) - \(\dfrac{1}{30}\)
A = 1 - \(\dfrac{1}{30}\)
A = \(\dfrac{29}{30}\)
17 tháng 4 2019
Chứng minh :
\(S=\frac{1}{5}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{25}+\frac{1}{101}+\frac{1}{102}+...+\frac{1}{105}< \frac{1}{2}\)
Nhóm các số hạng:
\(S=\frac{1}{5}+\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{25}\right)+\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{105}\right)< \frac{1}{5}+\frac{5}{21}+\frac{5}{101}< \frac{1}{5}+\frac{5}{20}+\frac{5}{100}=\frac{1}{2}.\)
XO
7 tháng 6 2019
1)
A = \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}\)
= \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{11.12}\)
= \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{12}\)
= \(\frac{1}{5}-\frac{1}{12}\)
= \(\frac{7}{60}\)
B = \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{99}\right)\)
= \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
= \(\frac{3.4.5.....100}{2.3.4....99}\)
= \(\frac{100}{2}=50\)
C = \(\frac{1}{4^{2-1}}+\frac{1}{6^{2-1}}+\frac{1}{8^{2-1}}...+\frac{1}{30^{2-1}}\)
= \(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{30}\)
= \(\frac{1}{2.2}+\frac{1}{2.3}+\frac{1}{2.4}+...+\frac{1}{2.15}\)
= \(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{2}.\frac{1}{4}+...+\frac{1}{2}.\frac{1}{15}\)
= \(\frac{1}{2}.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{15}\right)\)
7 tháng 6 2019
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{5}+\left(\frac{1}{6}-\frac{1}{6}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+\left(\frac{1}{10}-\frac{1}{10}\right)+\left(\frac{1}{11}-\frac{1}{11}\right)-\frac{1}{12}\)
\(A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
~ Hok tốt ~
27 tháng 2 2021
`1/3+ -1/4+1/5+ -1/6+1/7+1/6+1/(-5)+1/4+ -1/3`
`=(1/3-1/3)+(1/4-1/4)+(1/5-1/5)+(1/6-1/6)+1/7`
`=0+0+0+0+1/7`
`=1/7`
