Tính:(0,2)^3.[(-1^5)^-2]^-1+[(5/6)^6:2^3]-(0,125):1/64
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 3 2024
\(\left(\dfrac{5}{7}-\dfrac{7}{7}\right)-\left[0,2-\left(-\dfrac{2}{7}-\dfrac{1}{10}\right)\right]\)
=\(-\dfrac{2}{7}-\left[\dfrac{1}{5}+\dfrac{2}{7}+\dfrac{1}{10}\right]\)
=\(-\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{10}\)
=\(\left(-\dfrac{2}{7}-\dfrac{2}{7}\right)-\left(\dfrac{1}{5}+\dfrac{1}{10}\right)\)
=\(-\dfrac{4}{7}-\left(\dfrac{2}{10}+\dfrac{1}{10}\right)\)
=\(-\dfrac{4}{7}-\dfrac{3}{10}\)
=\(-\dfrac{40}{70}-\dfrac{21}{70}\)
=\(-\dfrac{61}{70}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 3 2024
(3 - \(\dfrac{1}{4}\) + \(\dfrac{2}{3}\)) - (5 - \(\dfrac{1}{3}\) - \(\dfrac{5}{6}\)) - (6 - \(\dfrac{7}{4}\) - \(\dfrac{3}{2}\))
= 3 - \(\dfrac{1}{4}\) + \(\dfrac{2}{3}\) - 5 + \(\dfrac{1}{3}\) + \(\dfrac{5}{6}\) - 6 + \(\dfrac{7}{4}\) + \(\dfrac{3}{2}\)
= (3 - 5 - 6) + ( \(\dfrac{7}{4}\) - \(\dfrac{1}{4}\)) + (\(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)) + \(\dfrac{3}{2}\) + \(\dfrac{5}{6}\)
= - 8 + \(\dfrac{3}{2}\) + 1 + \(\dfrac{3}{2}\) + \(\dfrac{5}{6}\)
= (- 8 + 1) + (\(\dfrac{3}{2}\) + \(\dfrac{3}{2}\)) + \(\dfrac{5}{6}\)
= -7 + 3 + \(\dfrac{5}{6}\)
= - 4 + \(\dfrac{5}{6}\)
= \(\dfrac{-19}{6}\)
31 tháng 10 2021
a, \(\dfrac{3}{5}.\left(-\dfrac{8}{3}\right)-\dfrac{3}{5}:\left(-\dfrac{3}{2}\right)=\dfrac{3}{5}.\left(-\dfrac{8}{3}\right)-\dfrac{3}{5}.\left(-\dfrac{2}{3}\right)==\dfrac{3}{5}\left(-\dfrac{8}{3}-\dfrac{2}{3}\right)=\dfrac{3}{5}.\left(-\dfrac{10}{3}\right)=-2\)
b, \(-\dfrac{5}{6}.\left(-\dfrac{12}{7}\right)-\left(-\dfrac{21}{15}\right)=-\dfrac{5}{6}.\left(-\dfrac{12}{7}\right)+\dfrac{7}{5}=\dfrac{10}{7}+\dfrac{7}{5}=\dfrac{50+49}{35}=\dfrac{99}{35}\)
31 tháng 10 2021
a: \(=\dfrac{3}{5}\cdot\left(-\dfrac{8}{3}+\dfrac{-2}{3}\right)=\dfrac{3}{5}\cdot\dfrac{-10}{3}=-2\)
c: \(=\left(0.125\right)^{650}\cdot8^{102}\)
\(=\left(0.125\cdot8\right)^{102}\cdot\left(0.125\right)^{548}\)
\(=\dfrac{1}{8^{548}}\)
PN
14 tháng 10 2017
a) \(1\dfrac{7}{20}:2,7+2,7:1,35+\left(0,4:2\dfrac{1}{2}\right).\left(4,2-1\dfrac{3}{40}\right)\)
\(=\dfrac{27}{20}:\dfrac{27}{10}+\dfrac{27}{10}:\dfrac{27}{10}+\left(\dfrac{2}{5}:\dfrac{5}{2}\right).\left(\dfrac{21}{5}-\dfrac{43}{40}\right)\)
\(=1+1+\dfrac{4}{25}.\dfrac{25}{8}\)
\(=2+\dfrac{1}{2}\)
\(=2\dfrac{1}{2}\)
PN
14 tháng 10 2017
b) \(\left(6\dfrac{3}{5}-3\dfrac{3}{14}\right).5\dfrac{5}{6}:\left(21-1.25\right):2,5\)
\(=\left(\dfrac{33}{5}-\dfrac{45}{14}\right).\dfrac{35}{6}:\left(-4\right):2,5\)
\(=\left(\dfrac{462}{60}-\dfrac{225}{60}\right).\dfrac{35}{6}.\dfrac{1}{-4}:\dfrac{5}{2}\)
\(=\dfrac{237}{60}.\dfrac{35}{6}.\dfrac{1}{-4}.\dfrac{2}{5}\)
\(=\dfrac{3.79.7.5.2}{5.14.3.2.\left(-4\right).5}\)
\(=\dfrac{79.7}{14.\left(-4\right).5}=\dfrac{553}{-280}\) (số xấu :v)
LN
1
17 tháng 4 2022
\(2\dfrac{1}{6}+\dfrac{5}{6}:3-0,125\times\left(-2\right)^2\)
\(=\dfrac{13}{6}+\dfrac{5}{18}-0,125\times4\)
\(=\dfrac{22}{9}-0,5\)
\(=\dfrac{35}{18}\)
9 tháng 7 2015
\(\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{8}-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}{\frac{3}{4}+\frac{1}{2}-\frac{3}{10}}\)\(=\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{3\left(\frac{1}{8}-\frac{1}{5}+\frac{1}{7}\right)}+\frac{60.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}\right)}{60.\left(\frac{3}{4}+\frac{1}{2}-\frac{3}{10}\right)}\)
\(=\frac{1}{3}+\frac{30+20-12}{45+30-18}\)\(=\frac{1}{3}+\frac{38}{57}=1\)
27 tháng 11 2016
\(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-0,2}{\frac{3}{4}+0,5-\frac{3}{10}}\)
= \(\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{8}-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{2}{4}+\frac{2}{6}-\frac{2}{10}}{\frac{3}{4}+\frac{3}{6}-\frac{3}{10}}\)
= \(\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{3.\left(\frac{1}{8}-\frac{1}{5}+\frac{1}{7}\right)}+\frac{2.\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{10}\right)}{3.\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{10}\right)}\)
= \(\frac{1}{3}+\frac{2}{3}\)
= \(\frac{3}{3}\)
= 1