9 h 30 ' + 21 ' 15s :5
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A = \(\frac{1}{5\times6}+\frac{3}{6\times9}+\frac{5}{9\times14}+\frac{7}{14\times21}+\frac{9}{21\times30}\)
A = \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{21}+\frac{1}{21}-\frac{1}{30}\)
A = \(1-\frac{1}{30}\)
A = \(\frac{29}{30}\)
Áp dụng công thức:
* Từ 0s đến 5s:
* Từ 5s đến 15s:
* Từ thời điểm lớn hơn 15s: Ta thấy chất điểm chuyển động đều (v = 6m/s không đổi) nên a3 = 0m/s2
\(\dfrac{5}{8}+\dfrac{7}{24}\) = \(\dfrac{15}{24}\) + \(\dfrac{7}{24}\) = \(\dfrac{22}{24}\) = \(\dfrac{11}{12}\)
\(\dfrac{6}{5}+\dfrac{9}{30}=\dfrac{36}{30}+\dfrac{9}{30}=\dfrac{45}{30}=\dfrac{3}{2}\)
\(\dfrac{7}{12}\) + \(\dfrac{5}{2}\) = \(\dfrac{7}{12}+\dfrac{30}{12}=\dfrac{37}{12}\)
\(\dfrac{9}{21}\) + \(\dfrac{6}{49}\) = \(\dfrac{3}{7}\) + \(\dfrac{6}{49}\) = \(\dfrac{21}{49}\) + \(\dfrac{6}{49}\) = \(\dfrac{27}{49}\)
a; \(\dfrac{9}{27}\) + \(\dfrac{7}{-49}\)
= \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\)
= \(\dfrac{7}{21}\) - \(\dfrac{3}{21}\)
= \(\dfrac{4}{21}\)
b; - \(\dfrac{12}{10}\) + \(\dfrac{-25}{30}\)
= - \(\dfrac{6}{5}\) - \(\dfrac{5}{6}\)
= -\(\dfrac{36}{30}\) - \(\dfrac{25}{30}\)
= \(\dfrac{-61}{30}\)
c; \(\dfrac{-20}{35}\) + \(\dfrac{-16}{-24}\)
= - \(\dfrac{4}{7}\) + \(\dfrac{2}{3}\)
= - \(\dfrac{12}{21}\) + \(\dfrac{14}{21}\)
= \(\dfrac{2}{21}\)
d; - \(\dfrac{21}{77}\) + \(\dfrac{10}{-35}\)
= - \(\dfrac{3}{11}\) - \(\dfrac{2}{7}\)
= - \(\dfrac{21}{77}\) - \(\dfrac{22}{77}\)
= - \(\dfrac{43}{77}\)
a.
= -30-15+21-30-15-9
= (-30 - 30)+(-15-15)+(21-9)
= -60 -30 + 12
= -78
b.
= (6-18)+(-5+4)+(-20+21)
= 12 + (-1) + 1
=12
c.
= 27.(19+31)
=27. 50
= 1350
d.
= (-25).(64+36)
= -25 . 100
= -2500
e. 18.36 + 72.15
= 18.36 + 4.18.15
=18.( 36 + 60)
= 18.96
= 1728
a, 5 ∉ ƯC (10, 12)
b, 3 ∈ ƯC (9, 15, 21)
c, 30 ∈ BC (5, 6, 10, 15)
d, 21 ∉ BC (7, 14, 18, 21)
a: 7/30=28/120
13/60=26/120
-9/40=-27/120
b: -3/5=-18/30
-5/6=-25/30
3=90/30
c: -4/7=-36/63
8/9=56/63
-10/21=-30/63
d: -3/20=-9/60
-11/-30=11/30=22/60
7/15=28/60
\(9h30'+21'15s\div5=9h30'+4'15s\)
\(=9h34'15s\)