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(-3/20).16/9+1 1/2
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
15 tháng 12 2023
a, (\(\dfrac{9}{10}\) - \(\dfrac{15}{16}\)) \(\times\) ( \(\dfrac{5}{12}\) - \(\dfrac{11}{15}\) - \(\dfrac{7}{20}\))
= (\(\dfrac{72}{80}\) - \(\dfrac{75}{80}\)) \(\times\) (\(\)\(\dfrac{25}{60}\) - \(\dfrac{44}{60}\) - \(\dfrac{21}{60}\))
= - \(\dfrac{3}{80}\) \(\times\) (- \(\dfrac{2}{3}\))
= \(\dfrac{1}{40}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
15 tháng 12 2023
b, (-1)3 + (- \(\dfrac{2}{3}\))2 : 2\(\dfrac{2}{3}\) + \(\dfrac{5}{6}\)
= -13 + \(\dfrac{4}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{3}{8}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{1}{6}\) + \(\dfrac{5}{6}\)
= -1 + 1
= 0
16 tháng 12 2023
1: \(78+22+18\)
\(=\left(78+22\right)+18\)
=100+18
=118
2: \(94+563+\left(106-563\right)-\left(-70\right)\)
\(=94+563+106-563+70\)
\(=\left(94+106\right)-\left(563-563\right)+70\)
=100-0+70
=170
3: \(25\cdot154-25+47\cdot25\)
\(=25\left(154-1+47\right)\)
\(=25\cdot200=5000\)
4: \(\left[5^{29}+5^{30}\left(16-11\right)\right]:5^{29}\)
\(=\left(5^{29}+5^{30}\cdot5\right):5^{29}\)
\(=\dfrac{5^{29}\cdot1+5^{29}\cdot5^2}{5^{29}}\)
\(=1+5^2=26\)
8: \(25\cdot2^3-\left(9-14\right)+\left(29-34+20\right)\)
\(=25\cdot8-\left(-5\right)+\left(-5\right)+20\)
\(=200+5-5+20\)
=220
6 tháng 1 2022
a, (-20) + 16 + (-34) + 20 = [(-20) + 20] + [16 + (-34)] = 0 + (-18) = (-18) b, 12 + 3.[39 - (5 - 2)2] = 12 + 3.[39 - 32] = 12 + 3.[39 - 9] = 12 + 3.30 = 12 + 90 = 102
H
2
7 tháng 5 2019
Trả lời :
a)\(-\frac{3}{5}+\frac{9}{16}+-\frac{2}{5}+\frac{7}{16}\)
=\(\left(-\frac{3}{5}+-\frac{2}{5}\right)+\left(\frac{9}{16}+\frac{7}{16}\right)\)
=\(-1+1\)
=\(1\)
b)\(5\frac{2}{3}-1\frac{1}{5}-2\frac{2}{3}\)
=\(\left(5-1-2\right)\left(\frac{2}{3}-\frac{1}{5}-\frac{2}{3}\right)\)
=\(-2\frac{1}{5}\)
Hok_Tốt
#Thiên_Hy
7 tháng 5 2019
a ) -3/5 + 9/16 + (-2/5 ) + 7/16
[ -3/5 + (-2/5) ] + (9/16 + 7/16 )
-1 + 1
0
b) 5 2/3 - 1 1/5 - 2 2/3
(5 2/3 - 2 2/3) - 1 1/5
3 - 6/1
-3
19 tháng 12 2020
a) Ta có: \(\left|5\cdot0.6+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=\left|3+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=3+\dfrac{2}{3}-\dfrac{1}{3}\)
\(=3+\dfrac{1}{3}=\dfrac{10}{3}\)
b) Ta có: \(\left(0.25-1\dfrac{1}{4}\right):5-\dfrac{1}{5}\cdot\left(-3\right)^2\)
\(=\left(\dfrac{1}{4}-\dfrac{5}{4}\right)\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{-4}{4}\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{1}{5}\cdot\left(-1-9\right)\)
\(=-10\cdot\dfrac{1}{5}=-2\)
c) Ta có: \(\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}:\dfrac{5}{7}\)
\(=\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\cdot\left(\dfrac{14}{17}+\dfrac{3}{17}\right)\)
\(=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)
d) Ta có: \(\dfrac{7}{16}+\dfrac{-9}{25}+\dfrac{9}{16}+\dfrac{-16}{25}\)
\(=\left(\dfrac{7}{16}+\dfrac{9}{16}\right)-\left(\dfrac{9}{25}+\dfrac{16}{25}\right)\)
\(=\dfrac{16}{16}-\dfrac{25}{25}\)
\(=1-1=0\)
e) Ta có: \(\dfrac{5}{6}+2\sqrt{\dfrac{4}{9}}\)
\(=\dfrac{5}{6}+2\cdot\dfrac{2}{3}\)
\(=\dfrac{5}{6}+\dfrac{4}{3}\)
\(=\dfrac{5}{6}+\dfrac{8}{6}=\dfrac{13}{6}\)
TN
1
5 tháng 12 2015
\(20\frac{15}{16}.16\frac{8}{9}-20\frac{15}{16}.12\frac{8}{9}\)
\(=20\frac{15}{16}.\left(16\frac{8}{9}-12\frac{8}{9}\right)\)
\(=20\frac{15}{16}.4\)
\(=83\frac{3}{4}\)
22 tháng 7 2021
a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)
\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)
\(=-1+1=0\)
b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)
\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)
=1-1+1=1
8 tháng 6 2018
b ) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
c ) Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
=> A < 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100= 1 - 1/100 = 99/100 < 1
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)< 1
NV
8 tháng 6 2018
b, \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\)\(\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
c,Ta thấy
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(.....\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
\(\left(\frac{-3}{20}\right).\frac{16}{9}+1\frac{1}{2}\)
\(=\frac{-4}{15}+\frac{3}{2}\)
\(=\frac{-8}{30}+\frac{45}{30}\)
\(=\frac{37}{30}\)