168:{11+[(-8)+3×(14-29)]}
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a)(-12)+(-9)+112+|-30|
=(-21)+112+30
= 91+ 30
=121
b) - 2015 + (-268)+(-201)+168+301
=(-2283)+(-33)+301
=(-2316)+301
= - 2015
c)160 :{|-17|+[3x2x5-(14+2x11:2x8)]}
= 160 :{17+[3x2x5-(14+22:16)]}
=160:{17+[ 3x2x5-15,375]
=160:{17+30-15,375}
=160:31,625
=5,05
a, ( - 168 ) + 72 . ( - 168 ) + ( - 168 ) . 27
= ( - 168 ) + ( - 12096 ) + ( - 4536 )
= - 12264 + - 4536
= - 16800
b, 22 . ( - 3 ) - ( 110 + 8 ) : ( - 3 )2
= 4 . ( - 3 ) - ( 1 + 8 ) : 9
= ( - 12 ) - 9 : 9
= ( - 12 ) - 1
= - 13
c, ( - 1075 ) - ( 29 - 1075 )
= ( - 1075 ) - ( - 1046 )
= - 29
d, ( - 9 ) + ( - 11 ) + 21 + ( - 1 )
= - 20 + 21 + ( - 1 )
= 1 + - 1
= 0
e, 30 + 12 + ( - 20 ) + ( - 12 ) - ( 30 - 20 ) + ( 12 - 12 )
= 42 + ( - 20 ) + ( - 12 ) - 10 + 0
= 22 + ( - 12 ) - 10 + 0
= 10 - 10 + 0
= 0 + 0
= 0
g, ( 13 - 135 + 49 ) - ( 13 + 49 )
= [( - 122 ) + 49 ] - 62
= ( - 73 ) - 62
= - 135
h, 35 - { 12 - [ ( - 14 ) + ( - 2 ) } ]
= 35 - { 12 - ( - 16 ) }
= 35 - 28
= 7
Bài 2:
a. x - 35 = ( - 12 ) - 3
x - 35 = - 15
x = - 15 + 35
x = 20
b, \(\frac{1}{4}\)+ \(\frac{1}{3}\): 3x = - 5
\(\frac{3}{12}+\frac{4}{12}\): 3x = - 5
\(\frac{7}{12}\): 3x = - 5
3x = \(\frac{7}{12}\): - 5
3x = \(\frac{-7}{60}\)
x = \(\frac{-7}{60}\): 3
x = \(\frac{-7}{180}\)
c,2x-1 = 8
2x-1 = 24
x = 4 + 1
x = 5
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
A . số số hạng =(302 - 5)/3 + 1 = 100
tổng dãy số = (302 + 5) .100/2 = 15350
Vậy tổng A= 15350
Làm tương tự như trên ta đc :
B= 5250
C=9210
P/s : đúng thì mk nhak, hì
D=8975
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)
=\(\dfrac{10}{11}.\dfrac{-1}{2}\)
=\(\dfrac{-5}{11}\)
b;
B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8
B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8
B = \(\dfrac{2}{7}\) - 8
B = \(\dfrac{2}{7}-\dfrac{56}{7}\)
B = - \(\dfrac{54}{7}\)
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{26.29}\)
\(=\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{26.29}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{26}-\frac{1}{29}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{29}\right)\)
\(=\frac{1}{3}.\frac{24}{145}\)
\(=\frac{8}{145}\)
Đặt A=\(\frac{1}{5\times8}+\frac{1}{8\times11}+.......+\frac{1}{26\times29}\)
Ta có: 3A=\(\frac{3}{5\times8}+\frac{3}{8\times11}+....+\frac{3}{26\times29}\)
\(\Rightarrow3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{26}-\frac{1}{29}\)
\(\Rightarrow3A=\frac{1}{5}-\frac{1}{29}\)
\(\Rightarrow A=\frac{24}{145}:3\)
\(\Rightarrow A=\frac{8}{145}\)
a)15/16:3/8x3/4
=15/16:6/16x12/16
=15/16x16/6x12/16
=15/6x12/16
=30/16=15/8
b)5/11x18/29-5/11x8/29+5/11x19/29
=5/11x(18/29-8/29+19/29)
=5/11x1=5/11
168 : {11 + [(-8) + 3 x (14 - 29)]}
= 168 :{11 + [- 8 + 3 x (-15)]}
= 168: {11 + [-8 - 45]}
= 168: {11 - 53}
= 168: {-42}
= - 4