(1/15)^2*15^2-2^5*5/2^7
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C
19 tháng 3 2023
`1)1/2:2/3 .... 2/3 : 1/2`
`=>1/2xx3/2 .... 2/3xx2`
`=>3/4 .... 4/3`
Vì `3/4 < 1` và `4/3>1`
`=>3/4<4/3`
__
`4/7:2/5 ... 4/7 : 3/5`
`=>4/7xx5/2....4/7xx5/3`
`=>20/14...20/21`
`=>10/7...20/21`
Vì `10/7>1` và `20/21<1`
`=>10/7>20/21`
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`4/15:4/7....2/5xx10/3`
`=>4/15xx7/4...20/15`
`=>7/15...20/15`
Vì `7<20` nên `7/15<20/15`
__
`5/6...15/18-11/18`
`=>5/6...4/18`
Ta có : MSC : `18`
`5/6 = 15/18`
Vì `15>4` nên `5/6 > 4/18`
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
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 + \(\dfrac{5}{6}\)
= \(\dfrac{-19}{6}\)
9 tháng 5 2019
\(\frac{3}{5}\times\frac{6}{7}+\frac{3}{5}:7+\frac{6}{5}\)
\(=\frac{3}{5}\times\frac{6}{7}+\frac{3}{5}\times\frac{1}{7}+\frac{6}{5}\)
\(=\frac{3}{5}\times\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{6}{5}\)
\(=\frac{3}{5}\times1+\frac{6}{5}\)
\(=\frac{3}{5}+\frac{6}{5}=\frac{9}{5}\)
~ Hok tốt ~
9 tháng 5 2019
\(\frac{29}{12}:\frac{1}{2}-\frac{5}{12}:\frac{1}{2}-\frac{1}{2}\)
\(=\left(\frac{29}{12}-\frac{5}{12}\right):\frac{1}{2}-\frac{1}{2}\)
\(=2:\frac{1}{2}-\frac{1}{2}\)
\(=4-\frac{1}{2}=\frac{7}{2}\)
~ Hok tốt ~
18 tháng 1 2024
a: \(\dfrac{16}{15}\cdot\dfrac{-5}{14}\cdot\dfrac{54}{24}\cdot\dfrac{56}{21}\)
\(=\dfrac{16}{24}\cdot\dfrac{-5}{15}\cdot\dfrac{54}{21}\cdot\dfrac{56}{14}\)
\(=\dfrac{2}{3}\cdot\dfrac{-1}{3}\cdot\dfrac{18}{7}\cdot4\)
\(=\dfrac{-2\cdot2\cdot4}{7}=-\dfrac{16}{7}\)
b: \(\dfrac{7}{3}\cdot\dfrac{-5}{2}\cdot\dfrac{15}{21}\cdot\dfrac{4}{-5}\)
\(=\dfrac{7}{3}\cdot\dfrac{5}{7}\cdot\dfrac{5}{2}\cdot\dfrac{4}{5}\)
\(=\dfrac{5}{3}\cdot\dfrac{4}{2}=\dfrac{5}{3}\cdot2=\dfrac{10}{3}\)
c: \(\left(-\dfrac{2}{5}+\dfrac{1}{3}\right)\left(\dfrac{3}{2}-\dfrac{3}{7}\right)\)
\(=\dfrac{-2\cdot5+3}{15}\cdot\dfrac{3\cdot7-3\cdot2}{14}\)
\(=\dfrac{-7}{15}\cdot\dfrac{15}{14}=\dfrac{-7}{14}=-\dfrac{1}{2}\)
d: \(\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\left(5-\dfrac{1}{4}\right)\)
\(=\dfrac{3-2}{6}\cdot\dfrac{20-1}{4}\)
\(=\dfrac{1}{6}\cdot\dfrac{19}{4}=\dfrac{19}{24}\)
e: \(\left(\dfrac{4}{5}\right)^2=\dfrac{4^2}{5^2}=\dfrac{16}{25}\)
5 tháng 3 2023
a: \(=\left(15-6-\dfrac{13}{18}\right):\dfrac{298}{27}-\dfrac{17}{8}:\dfrac{51}{40}\)
\(=\dfrac{149}{18}\cdot\dfrac{27}{298}-\dfrac{5}{3}=\dfrac{3}{2}-\dfrac{5}{3}=\dfrac{9-10}{6}=\dfrac{-1}{6}\)
b: \(=\dfrac{-16}{5}\cdot\dfrac{-15}{64}+\dfrac{-22}{15}:\dfrac{11}{2}\)
\(=\dfrac{3}{4}-\dfrac{4}{15}=\dfrac{29}{60}\)
c: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=\dfrac{-7}{9}+\dfrac{7}{9}+5=5\)
d: \(=\dfrac{1}{2}\cdot\dfrac{4}{3}\cdot10\cdot\dfrac{1}{5}\cdot\dfrac{3}{4}=1\)
e: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}+\dfrac{-23}{4}=\dfrac{204}{25}\)
\(=\left(\dfrac{1}{15}\cdot15\right)^2-\dfrac{2^5\cdot5}{2^7}=1-\dfrac{5}{4}=-\dfrac{1}{4}\)