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M=1/3+1/6+1/12+1/24+1/48+1/96+1/192
N=7/3+7/15+7/35+7/63+7/99+7/143+7/195
Ai làm nhanh mk tik cho nha
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Những câu hỏi liên quan
NT
Nguyễn Tuấn Tú
CTVHS
26 tháng 2 2023
2/3 + 7/1 = 2/3 + 7 = 2/3 + 21/3 = 23/3
25/48 + 11/24 = 25/48 + 22/48 = 47/48
5/7 + 3/8 = 40/56 + 21/56 = 61/56
15/24 + 12/6 = 5/8 + 2 = 5/8 + 16/8 = 21/8
5/6 + 4/3 = 5/6 + 8/6 = 13/6
3/8 + 7/12 = 9/24 + 14/24 = 23/24
5 tháng 6 2023
\(\dfrac{7}{19}x\dfrac{8}{23}+\dfrac{7}{19}x\dfrac{15}{23}+1\dfrac{7}{19}\)
= \(\dfrac{7}{19}x\left(\dfrac{8}{23}+\dfrac{15}{23}\right)+1+\dfrac{7}{19}\)
=\(\dfrac{7}{19}x1+1+\dfrac{7}{19}\)
= \(\dfrac{7}{19}+1+\dfrac{7}{19}=1\dfrac{14}{19}\) = \(\dfrac{33}{19}\)
\(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{49}{32}+\dfrac{1}{4}+\dfrac{3}{21}-\dfrac{17}{32}\)
= \(\dfrac{3}{4}+\dfrac{6}{7}+\dfrac{49}{32}+\dfrac{1}{4}+\dfrac{1}{7}-\dfrac{17}{32}\)
= \(\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{6}{7}+\dfrac{1}{7}\right)+\left(\dfrac{49}{32}-\dfrac{17}{32}\right)\)
= 1 + 1 + 1 = 3
\(\dfrac{8}{9}x\dfrac{15}{16}x\dfrac{24}{25}x\dfrac{35}{36}x\dfrac{48}{49}x\dfrac{63}{64}\)
= \(\dfrac{3}{4}\) *Câu này bạn tự sử dụng gạch nhé!
5 tháng 6 2023
`1,`
`a,`
`7/19 \times 8/23 + 7/19 \times 15/23 + 1 7/19`
`= 7/19 \times 8/23 + 7/19 \times 15/23 + 1 + 7/19`
`= 7/19 \times (8/23 + 15/23 + 1) + 1`
`= 7/19 \times 2 + 1`
`=14/19 + 1`
`= 33/19`
`b,`
`75/100 + 18/21 + 49/32 + 1/4 + 3/21 - 17/32`
`= 75/100 + (18/21 + 3/21) + (49/32 - 17/32) + 1/4`
`= 0,75 + 1 + 1 + 0,25`
`= (0,75 + 0,25) + 1 + 1`
`= 1+1+1=3`
`c,`
`8/9 \times 15/16 \times 24/25 \times 35/36 \times 48/49 \times 63/64`
`=` \(\dfrac{2\times3}{3\times3}\times\dfrac{3\times5}{4\times4}\times\dfrac{3\times4\times2}{5\times5}\times\dfrac{5\times7}{6\times6}\times\dfrac{6\times8}{7\times7}\times\dfrac{7\times9}{8\times8}\)
`= 3/4` (bạn sử dụng gạch, rút gọn các số là được nhé).
TA
21 tháng 2 2018
a)\(\frac{7}{1}.\frac{5}{9}.\frac{11}{35}=\frac{7.5.11}{9.5.7}=\frac{11}{9}\)
b) \(\frac{3}{5}.\frac{4}{23}.\frac{5}{6}=\frac{3.2.2.5}{5.23.3.2}=\frac{2}{23}\)
c) \(\frac{5}{2}.\frac{7}{4}.\frac{2}{7}=\frac{5.7.2}{2.7.4}=\frac{5}{4}\)
d) \(\frac{19}{23}.\frac{12}{17}.\frac{85}{24}=\frac{19.12.5.17}{23.17.12.2}=\frac{19.5}{23.2}=\frac{95}{46}\)
S
21 tháng 2 2018
Tính
A. \(\frac{7}{1}x\frac{5}{9}x\frac{11}{35}=\frac{7x5x11}{1x9x35}=\frac{11}{9}\)
B. \(\frac{3}{5}x\frac{4}{23}x\frac{5}{6}=\frac{3x4x5}{5x23x6}=\frac{2}{23}\)
C. \(\frac{5}{2}x\frac{7}{4}x\frac{2}{7}=\frac{5x7x2}{2x4x7}=\frac{35}{28}\)
D. \(\frac{19}{23}x\frac{12}{17}x\frac{85}{24}=\frac{19x12x85}{23x17x24}=\frac{4845}{2346}\)
13 tháng 2 2018
Ta có :
a) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\)\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)
\(=\)\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\)\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\)\(\frac{1}{2}.\frac{4}{15}\)
\(=\)\(\frac{2}{15}\)
13 tháng 2 2018
Ta có :
\(c)\)\(\frac{1}{1000}+\frac{13}{1000}+\frac{25}{1000}+\frac{37}{1000}+...+\frac{229}{1000}\)
\(=\)\(\frac{1+13+25+37+...+229}{1000}\)
Xét tổng \(1+13+25+37+...+229\):
Số số hạng : \(\left(229-1\right):12+1=20\) ( số hạng )
Tổng : \(\frac{\left(229+1\right).20}{2}=2300\)
Do đó :
\(\frac{1+13+25+37+...+229}{1000}=\frac{2300}{1000}=\frac{23}{10}\)
MH
4
N
23 tháng 6 2018
=7/2.(2/1.3+2/3.5+....+2/13.15)
=7/2.(1/1-1/3+1/3-1/5+....+1/13-1/15)
=7/2.(1-1/15)
=7/2.14/15=7.14/2.25
=7.7/1/15=49/15
14 tháng 3 2020
a,
A=1−3−5−7−9−...−97−99a)A=1−3−5−7−9−...−97−99
=1−(3+5+7+...+99)=1−(3+5+7+...+99)
=1−(99+3).[(99−3):2+1]2=1−(99+3).[(99−3):2+1]2
=1−2499=−2498=1−2499=−2498
b)B=1+3−5−7+9+...+97−99b)B=1+3−5−7+9+...+97−99
=(−8)+(−8)+(−8)+...+(−8)+97−99=(−8)+(−8)+(−8)+...+(−8)+97−99
=(−8).12+(−2)=−98=(−8).12+(−2)=−98
c)C=1−3−5+7+9−11−13+15+...+97−99c)C=1−3−5+7+9−11−13+15+...+97−99
=0+0+0+0+0+...+0−99=0+0+0+0+0+...+0−99
=−99
18 tháng 8 2017
Bài 1:
\(A=\frac{5}{3.6}+\frac{5}{6.9}+....+\frac{5}{96.99}\)
\(\Rightarrow\frac{3}{5}A=\frac{3}{3.6}+\frac{3}{6.9}+....+\frac{3}{96.99}\)
\(\Rightarrow\frac{3}{5}A=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{96}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\Rightarrow A=\frac{32}{99}\div\frac{3}{5}=\frac{160}{297}\)
Bái 2:
\(B=\frac{2}{3.7}+\frac{2}{7.11}+...+\frac{2}{99.103}\)
\(\Rightarrow2B=\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{99.103}\)
\(\Rightarrow2B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{99}-\frac{1}{103}\)
\(=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)
\(\Rightarrow B=\frac{100}{309}\div2=\frac{50}{309}\)
18 tháng 8 2017
Bài 1:
Ta có:
\(\frac{5}{n.\left(n+3\right)}=\frac{5}{3}.\frac{3}{n.\left(n+3\right)}=\frac{5}{3}.\frac{\left(n+3\right)-n}{n.\left(n+3\right)}=\frac{5}{3}.\left[\frac{n+3}{n.\left(n+3\right)}-\frac{n}{n\left(n+3\right)}\right]\)\(=\frac{5}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)\)
\(\frac{5}{3.6}+\frac{5}{6.9}+\frac{5}{9.12}+...+\frac{5}{96.99}=\frac{5}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{96}-\frac{1}{99}\right)\)
3 tháng 2 2022
a) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
=\(1-\dfrac{1}{6}\)=\(\dfrac{5}{6}\)
b) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
=\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
=\(\dfrac{1.2}{3.5.2}+\dfrac{1.2}{5.7.2}+\dfrac{1.2}{7.9.2}+\dfrac{1.2}{9.11.2}+\dfrac{1.2}{11.13.2}\)
=\(\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\).
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)=\(\dfrac{1}{2}.\dfrac{10}{39}\)=\(\dfrac{5}{39}\).
c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
=\(1-\dfrac{1}{8}=\dfrac{7}{8}\).
d) \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4}{2^5}+\dfrac{2^3}{2^5}+\dfrac{2^2}{2^5}+\dfrac{2}{2^5}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4+2^3+2^2+2+1}{2^5}\)=\(\dfrac{2^5-1}{2^5}=\dfrac{31}{32}\).
e) \(\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}=\dfrac{7^{99}+7^{98}+7^{97}+...+7+1}{7^{100}}=\dfrac{\dfrac{7^{100}-1}{6}}{7^{100}}=\dfrac{7^{100}-1}{6.7^{100}}\)
N=7/2(2/1.3+....+2/13.15)
N=7/2.(1/1-1/3+.....+1/13-1/15)
N=7/2.(1-1/15)
N=7/2.(14/15)
N=7.14/2.15
Bạn giải M luôn dc ko?