Bài 1: (1,0 điểm) Tính bằng cách hợp lý nếu có thể:
a) 2011 : { 639 : [ 316 – ( 78 + 25 )] : 3 }
b) (-21) + -50+ (-29) – -2016
giúp mik vs
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a) 2011 : { 639 : [ 316 – ( 78 + 25 )] : 3 }
= 2011 : { 639 : [ 316 – 103 ] : 3}
= 2011 : ( 639 : 213 : 3 ) = 2011 : (3 : 3 ) = 2011 : 1 = 2011
a) (-21) + |-50| + (-29) – |-2016|
= (-21) + 50 + (-29) – 2016
= [(-21) + (-29) + 50] – 2016 = ( -50 +50 ) – 2016
= 0 – 2016 = - 2016 .
a) (-124) + 24 = -100
b) 37. 78 + 37. 22
= 37 . ( 78 + 22 )
= 37 . 100
= 3700
a) (-21) + |-50| + (-29) – |-2016|
= (-21) + 50 + (-29) – 2016
= [(-21) + (-29) + 50] – 2016 = ( -50 +50 ) – 2016
= 0 – 2016 = - 2016 .
b) 36 : 3 2 + 3 2 . 2 3 - 15 0
= 36 : 9 + 9 . 8 – 1 = 4 + 72 – 1 = 76 – 1 = 75
c) ( 5 103 – 5 102 – 5 101 ) : ( 5 99 . 26 – 5 99 )
= ( 5 103 – 5 102 – 5 101 ) : 5 99 . ( 26 – 1 )
= ( 5 103 – 5 102 – 5 101 ) : ( 5 99 . 25 )
= 5 101 ( 5 2 – 5 1 – 5 0 ) : ( 5 99 . 5 2 )
= ( 5 101 . 19 ) : 5 101 = 19
a) 2011 : { 639 : [ 316 – ( 78 + 25 )] : 3 }
= 2011 : { 639 : [ 316 – 103 ] : 3}
= 2011 : ( 639 : 213 : 3 ) = 2011 : (3 : 3 ) = 2011 : 1 = 2011
b) ( 3x – 2 3 ) . 7 = 7 4
3x – 8 = 7 4 : 7
3x – 8 = 7 3
3x – 8 = 343
3x = 343 + 8
3x = 351
x = 351 : 3 = 117
c) (8705 + 5235) – 5x = 3885
13940 – 5x = 3885
5x = 13940 – 3885
5x = 10055
x = 10055 : 5 = 2011
a) 2011 : { 639 : [ 316 – ( 78 + 25 )] : 3 }
= 2011 : { 639 : [ 316 – 103 ] : 3}
= 2011 : ( 639 : 213 : 3 ) = 2011 : (3 : 3 ) = 2011 : 1 = 2011
b) ( 3x – 23) . 7 = 74
3x – 8 = 74 : 7
3x – 8 = 73
3x – 8 = 343
3x = 343 + 8
3x = 351
x = 351 : 3 = 117
c) (8705 + 5235) – 5x = 3885
13940 – 5x = 3885
5x = 13940 – 3885
5x = 10055
x = 10055 : 5 = 2011
Trl:
a) \(2011:\left\{639:\left[316-\left(78+25\right)\right]:3\right\}\)
\(=2011:\left\{639:\left[316-103\right]:3\right\}\)
\(=2011:\left\{639:213:3\right\}\)
\(=2011:1\)
\(=2011\)
b) \(\left(3x-23\right).7=74\)
\(\Rightarrow3x-23=74:7\)
\(\Rightarrow3x-23=10,5\)
\(\Rightarrow3x=10,5+23\)
\(\Rightarrow3x=33,5\)
\(\Rightarrow x=33,5:3\)
\(\Rightarrow11,1\)( Câu này sai đề nha )
c) \(\left(8705+5235\right)-5x=3885\)
\(\Rightarrow13940-5x=3885\)
\(\Rightarrow5x=10055\)
\(\Rightarrow x=10055:5\)
\(\Rightarrow x=2011\)
\(a)\)
\(\left(-31\right)+\left(50-19\right)-\left(150-31\right)\)
\(=\left(-31\right)+50-19-150+31\)
\(=\left(-150\right)-19\)
\(=-169\)
\(b)\)
\(25.\left(45-17\right)-45.\left(25-17\right)\)
\(=25.45-25.17-45.25+45.17\)
\(=0\)
\(c)\)
\(\frac{-1}{12}+\frac{4}{3}=\frac{5}{4}\)
\(d)\)
\(3+\frac{-5}{20}+\frac{30}{75}+\frac{-7}{4}\)
\(=\left(\frac{3}{5}+\frac{30}{75}\right)-\left(\frac{5}{20}+\frac{7}{4}\right)\)
\(=1-2\)
\(=-1\)
a) (-31)+(50-19)-(150-31)
= (-31)+50+(-19)-150+(-31)
= (-31)+50-150+(-19)-(-31)
= (-31)+(-100)+12
= -119
b) 25(45-17)-45(25-17)
= 25.45-25.17-45.25-45.17
= 25(45-45)-25(17-17)
= 0
c) -1/12 + 4/3
= -1/12 + 16/12
= 15/12
= 5/4
d) 3/5+(-5)/20+30/75+(-7)/4
= 45/75+30/75+(-5)/20+(-35)/20
= 1+(-2)
= -1