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NT
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
28 tháng 6 2016
\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}.\)
\(B< \frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{8-7}{7.8}\)
\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)
\(B< 1-\frac{1}{8}< 1\)
toi nay mk se lay y kien cua cac ban , giup mk nha
CV
12 tháng 7 2017
b1
a) \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{5}-\dfrac{1}{10}\)
\(=\dfrac{2}{10}-\dfrac{1}{10}\)
\(=\dfrac{1}{10}\)
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{1}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
c) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)
\(=\dfrac{1}{3}-\dfrac{1}{11}\)
\(=\dfrac{8}{33}\)
d) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=\dfrac{1}{3}-\dfrac{1}{101}\)
\(=\dfrac{98}{303}\)
QA
6 tháng 3 2020
Tính:
\(\frac{2}{3}+\frac{3}{4}=\frac{8}{12}+\frac{9}{12}=\frac{17}{12}\)
\(\frac{1}{5}+\frac{3}{2}=\frac{2}{10}+\frac{15}{10}=\frac{17}{10}\)
\(\frac{6}{5}-\frac{2}{3}=\frac{18}{15}-\frac{10}{15}=\frac{8}{15}\)
\(\frac{7}{2}-\frac{3}{4}=\frac{14}{4}-\frac{3}{4}=\frac{11}{4}\)
\(1\frac{2}{5}+\frac{7}{2}=\frac{7}{5}+\frac{7}{2}=\frac{14}{10}+\frac{35}{10}=\frac{49}{10}\)
\(2\frac{1}{8}-\frac{5}{6}=\frac{17}{8}-\frac{5}{6}=\frac{51}{24}-\frac{20}{24}=\frac{31}{24}\)
\(0,75+2\frac{1}{3}=\frac{3}{4}+\frac{7}{3}=\frac{9}{12}+\frac{28}{12}=\frac{37}{12}\)
\(1-\frac{1}{4}-0,25=1-\frac{1}{4}-\frac{1}{4}=\frac{4}{4}-\frac{1}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\)
\(\frac{11}{2}-\frac{5}{3}+\frac{7}{4}=\frac{66}{12}-\frac{20}{12}+\frac{21}{12}=\frac{67}{12}\)
\(\frac{5}{6}+1\frac{2}{3}-\frac{1}{4}=\frac{5}{6}+\frac{5}{3}-\frac{1}{4}=\frac{10}{12}+\frac{20}{12}-\frac{3}{12}=\frac{27}{12}=\frac{9}{4}\)
\(0,5+2\frac{3}{5}-\frac{5}{4}=\frac{1}{2}+\frac{13}{5}-\frac{5}{4}=\frac{10}{20}+\frac{52}{20}-\frac{25}{20}=\frac{37}{20}\)
\(3\frac{1}{2}-\frac{5}{6}+1,2=\frac{7}{2}-\frac{5}{6}+\frac{6}{5}=\frac{105}{30}-\frac{25}{30}+\frac{36}{30}=\frac{116}{30}=\frac{58}{15}\)
\(\frac{1}{3}+2\frac{1}{4}=\frac{1}{3}+\frac{9}{4}=\frac{4}{12}+\frac{27}{12}=\frac{31}{12}\)
\(\frac{7}{2}+\frac{3}{4}+\frac{7}{12}=\frac{42}{12}+\frac{9}{12}+\frac{7}{12}=\frac{58}{12}=\frac{29}{6}\)
\(2-\frac{4}{9}-\frac{1}{6}=\frac{36}{18}-\frac{8}{18}-\frac{3}{18}=\frac{25}{18}\)
Chúc bạn hok tốt !!! ღℓїŋɦ✿şʉŋღ
4 tháng 8 2023
a: \(12\dfrac{1}{3}-\left(3\dfrac{3}{4}+4\dfrac{3}{4}\right)\)
\(=\dfrac{37}{3}-3-4-\dfrac{3}{2}\)
\(=\dfrac{74-9}{6}-7=\dfrac{65}{6}-7=\dfrac{65-42}{7}=\dfrac{23}{7}\)
b: \(3\dfrac{5}{6}+2\dfrac{1}{6}\cdot6\)
\(=3+\dfrac{5}{6}+\dfrac{13}{6}\cdot6\)
\(=16+\dfrac{5}{6}=\dfrac{101}{6}\)
c: \(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)
\(=3+\dfrac{1}{2}+4+\dfrac{5}{7}-5-\dfrac{5}{14}\)
\(=2+\dfrac{7+10-5}{14}=2+\dfrac{12}{14}=2+\dfrac{6}{7}=\dfrac{20}{7}\)
d: \(=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}=\dfrac{9}{2}+\dfrac{1}{11}=\dfrac{99+2}{22}=\dfrac{101}{22}\)
Q
2 tháng 4 2017
Bài 1: Tính
\(\text{1)}\) \(\dfrac{5}{8}.\dfrac{7}{30}-\dfrac{5}{2}.\dfrac{1}{8}\)
\(=\dfrac{5}{8}.\dfrac{7}{30}-\dfrac{5}{8}.\dfrac{1}{2}\)
\(=\dfrac{5}{8}.\left(\dfrac{7}{30}-\dfrac{1}{2}\right)\)
\(=\dfrac{5}{8}.\dfrac{-4}{15}\)
\(=\dfrac{-1}{6}\)
\(\text{2)}\) \(\dfrac{21}{10}.\dfrac{3}{4}-\dfrac{21}{10}-\dfrac{3}{4}\)
\(=\dfrac{63}{40}-\dfrac{21}{10}-\dfrac{3}{4}\)
\(=\dfrac{-21}{40}-\dfrac{3}{4}\)
\(=\dfrac{-51}{40}\)
\(\text{3)}\) \(\dfrac{-4}{11}:\dfrac{-6}{11}\)
\(=\dfrac{-4}{11}.\dfrac{11}{-6}\)
\(=\dfrac{4}{6}\)
\(\text{4)}\) \(\dfrac{2}{7}.\dfrac{14}{3}-1\)
\(=\dfrac{4}{3}-1\)
\(=\dfrac{1}{3}\)
\(\text{5)}\) \(\dfrac{4}{7}:\left(\dfrac{1}{5}.\dfrac{4}{7}\right)\)
\(=\dfrac{4}{7}:\dfrac{1}{5}:\dfrac{4}{7}\)
\(=1:\dfrac{1}{5}\)
\(=5\)
\(\text{6)}\) \(\dfrac{12}{7}.\dfrac{7}{4}+\dfrac{35}{11}:\dfrac{245}{121}\)
\(=3+\dfrac{35}{11}.\dfrac{121}{245}\)
\(=3+\dfrac{11}{7}\)
\(=3\dfrac{11}{7}=\dfrac{32}{7}\)
\(\text{7)}\) \(\left(\dfrac{4}{3}+\dfrac{8}{3}\right).\left(\dfrac{7}{4}-\dfrac{6}{4}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)
\(=4.\left(\dfrac{7}{4}-\dfrac{6}{4}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)
\(=4.\dfrac{1}{4}:\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)
\(=4.\dfrac{1}{4}:\dfrac{19}{5}\)
\(=1:\dfrac{19}{5}\)
\(=\dfrac{5}{19}\)
\(\text{8)}\) \(\left(\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{\dfrac{1}{9}}{\dfrac{1}{9}}\right):\left(\dfrac{2}{3}+\dfrac{\dfrac{7}{15}}{\dfrac{2}{5}}-\dfrac{1}{6}\right)\)
\(=\left(0+1\right):\left(\dfrac{2}{3}+\dfrac{7}{15}:\dfrac{2}{5}-\dfrac{1}{6}\right)\)
\(=1:\left(\dfrac{2}{3}+\dfrac{7}{6}-\dfrac{1}{6}\right)\)
\(=1:\left(\dfrac{2}{3}+1\right)\)
\(=1:\dfrac{5}{3}\)
\(=\dfrac{3}{5}\)
\(\text{9)}\)
\(\left[\left(\dfrac{2}{193}-\dfrac{3}{389}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\dfrac{199}{75077}.\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\dfrac{199}{6613}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\left[\dfrac{3}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\left[\dfrac{3}{50}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\dfrac{114}{25}\)
\(=\dfrac{330875}{1507764}\)
C
11 tháng 2 2023
`#lv`
`-2/5+(-7/10)-9/6`
`=-2/5-7/10-9/6`
`=-4/10-7/10-9/6`
`=-11/10-3/2`
`=-11/10-15/10`
`=-26/10`
`=-13/5`
__
`1/2+(-2/5)-(-2/3)`
`=1/2-2/5+2/3`
`=5/10-4/10+2/3`
`=1/10+2/3`
`=3/30+20/30`
`=23/30`
__
`5/3-3/4+7/6`
`=20/12-9/12+14/12`
`=11/12+14/12`
`=25/12`
__
`-1/5+5/3-3/2`
`=-3/15+25/15-3/2`
`=22/15-3/2`
`=44/30-45/30`
`=-1/30`
__
`1/4-7/8+(-9/10)`
`=2/8-7/8-9/10`
`=-5/8-9/10`
`=-25/40-36/40`
`=-61/40`
__
`5/4+1/2+(-7/12)`
`=15/12+6/12-7/12`
`=21/12-7/12`
`=14/12`
`=7/6`
__
`-5/8-1/3+(-7/6)`
`=-5/8-1/3-7/6`
`=-15/24-8/24-28/24`
`=-52/24`
`=-13/6`
__
`-3/4-7/10+(-5/6)`
`=-3/4-7/10-5/6`
`=-45/60-42/60-50/60`
`=-137/60`
KH
11 tháng 2 2023
Bài làm
a, (-2/5) + (-7/10) - 9/6 =(-12/30) + (-21/30) - 45/30=(-33/30) - 45/30 = (-78/30)
b,1/2 + (-2/5) - (-2/3) = 15/30 + (-12/30) - (-20/30) = 3/30 - (-20/30) = 23/30
c,5/3 - 3/4 + 7/6 = 40/24 - 18/24 + 28/24 = 22/24 + 28/24 = 50/24 = 25/12
d, (-1/5) + 5/3 - 3/2 = (-6/30) + 50/30 - 45/30 = 44/30 - 45/30 = (-1/30)
e, 1/4 - 7/8 + (-9/10) = 10/40 - 35/40 - 36/40 = (-25/40) - 36/40 = (-61/40)
f, 5/4 +1/2 + (-7/12) = 15/12 + 6/12 - 7/12 = 21/12 - 7/12 = 14/12 = 7/6
g,(-5/8) -1/3 + (-7/6) = (-15/24) - 8/24 - 28/24 = (-23/24) - 28/24 = (-51/24)
k,(-3/4) - 7/10 + (-5/6) = (-45/60) - 42/60 - 50/60 = (-87/60) - 50/60 = (-137/60)
chúc bn học tốt nha
sai thì mn góp ý giúp mk
2 tháng 4 2017
1) \(\frac{5}{8}.\frac{7}{3}-\frac{5}{2}.\frac{1}{8}=\frac{5}{8}.\frac{7}{3}-\frac{5}{8}.\frac{1}{2}=\frac{5}{8}\left(\frac{7}{3}-\frac{1}{2}\right)=\frac{5}{8}.\frac{11}{6}=\frac{55}{48}\)
2) \(\frac{21}{10}.\frac{3}{4}-\frac{21}{10}.\frac{3}{4}=\frac{21}{10}\left(\frac{3}{4}-\frac{3}{4}\right)=\frac{21}{10}.0=0\)
3) \(\frac{-4}{11}:\frac{-6}{11}=\frac{-4}{11}.\frac{-11}{6}=\frac{-4.\left(-11\right)}{11.6}=\frac{-4.\left(-1\right)}{1.6}=\frac{4}{6}=\frac{2}{3}\)
4)\(\frac{2}{7}.\frac{14}{3}-1=\frac{2.14}{7.3}-1=\frac{2.2}{1.3}-1=\frac{4}{3}-1=\frac{1}{3}\)
5)\(\frac{4}{7}:\left(\frac{1}{5}.\frac{4}{7}\right)=\frac{4}{7}:\frac{4}{35}=\frac{4}{7}.\frac{35}{4}=\frac{4.35}{7.4}=\frac{1.5}{1.1}=5\)
6) \(\frac{12}{7}.\frac{7}{4}+\frac{35}{11}:\frac{245}{121}=\frac{12.7}{7.4}+\frac{35}{11}.\frac{121}{245}=\frac{3.1}{1.1}+\frac{35}{11}.\frac{121}{245}=3+\frac{35}{11}.\frac{121}{245}=3+\frac{35.121}{11.245}=\frac{1.11}{1.7}=\frac{11}{7}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{4}< \frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{9}< \frac{1}{2.3}\)
\(\frac{1}{4^2}=\frac{1}{16}< \frac{1}{3.4}\)
......
\(\frac{1}{8^2}=\frac{1}{64}< \frac{1}{7.8}\)
Ta có : \(VP< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)
\(=1-\frac{1}{8}=\frac{7}{8}\)
Mà \(\frac{7}{8}< 1\)Nên \(B< 1\left(đpcm\right)\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
...
\(\frac{1}{8^2}=\frac{1}{8\cdot8}< \frac{1}{7\cdot8}\)
=> \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
=> \(B< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)
=> \(B< \frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)
Lại có : \(\frac{7}{8}< 1\)
=> \(B< \frac{7}{8}< 1\Rightarrow B< 1\left(đpcm\right)\)