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17 tháng 3 2022
1)\(\dfrac{-5}{2}:\dfrac{1}{4}\) = \(\dfrac{-5}{2}\) x \(\dfrac{4}{1}\) = \(\dfrac{-20}{2}\)
15 tháng 4 2020
a) -( -35 + 7 - 13 ) + ( 2 - 35 ) - ( 13 - 7 )
= 35 - 7 + 13 + 2 - 35 - 13 + 7
= ( 35 - 35 ) + ( -7 + 7 ) + ( 13 - 13 ) + 2
= 0 + 0 + 0 + 2
= 2
b) ( 70 + 8 - 35 ) - ( 8 - 37 - 10 ) - 70
= 70 + 8 - 35 - 8 + 37 + 10 - 70
= ( 70 - 70 ) + ( 8 - 8 ) + ( -35 + 37 ) + 10
= 0 + 0 + 2 + 10
= 12
c) 5 - ( 14 + 4 - 52 ) - ( 52 - 14 ) + 4
= 5 - 14 - 4 + 52 - 52 + 14 + 4
= 5 - ( 14 + 14 ) + ( 52 - 52 ) + ( -14 + 14 )
= 5 - 14 - 14 + 0 + 0
= 5 - ( 14 - 14 )
= 5 - 0
= 5
17 tháng 2 2020
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)
CM
21 tháng 4 2018
x + 22 + (–14 ) + 52
= x + 22 + 52 – 14
= x + (22 + 52) – 14
= x + 74 – 14
= x + 60
12 tháng 4 2017
a) \(\dfrac{298}{719}:\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{3}\right)-\dfrac{2011}{2012}=\dfrac{298}{719}:\left(\dfrac{3}{12}+\dfrac{1}{12}+\dfrac{4}{12}\right)-\dfrac{2011}{2012}=\dfrac{298}{719}:\left(\dfrac{3+1+4}{12}\right)-\dfrac{2011}{2012}=\dfrac{298}{719}:\dfrac{2}{3}-\dfrac{2011}{2012}=\dfrac{298}{719}\cdot\dfrac{3}{2}-\dfrac{2011}{2012}=\dfrac{149.3}{719.1}-\dfrac{2011}{2012}=\dfrac{447}{719}-\dfrac{2011}{2012}=\dfrac{889364}{1446628}-\dfrac{1445909}{1446628}=\dfrac{889364-1445909}{1446628}=-\dfrac{556545}{1446628}.\)b)\(\dfrac{27\cdot18+27+103-120\cdot27}{15\cdot33+33\cdot12}=\dfrac{27\left(18+103-120\right)}{33\left(15+12\right)}=\dfrac{27\cdot1}{33\cdot27}=\dfrac{1\cdot1}{33\cdot1}=\dfrac{1}{33}\)
NK
2
20 tháng 12 2016
a)=x+22-14+52 b)=(-90)-p-10+100
=x+22+52-14 =(-90)-10-p+100
=x+74-14 =(-100)-p+100
=x+60 =(-100)+100-p
=0-p=-p
20 tháng 5 2018
a) x + 22 + (-14) + 52 = x + [ 22 + 52 ] + (-14) = x + 74 + (-14)
= x + [ 74 + (-14) ] = x + 60
b) (-90) - (p + 10) + 100 = (-90) - p - 10 + 100 = - p - 90 - 10 + 100
= -p - (90 + 10) + 100 = -p - 100 + 100 = -p + [ (-100) + 100] = -p
NT
1
DP
12 tháng 8 2017
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=2\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2\cdot\left(1-\frac{1}{20}\right)\)
\(=2\cdot\frac{19}{20}=\frac{19}{10}\)