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Bài 1:
A,
(100+121+144)÷(169+196)
=365:365=1
B
,1.2.3...7.8.(9-1-8)
=1.2.3...7.8.0=0
C,
3^2.2^4.2^32/11.2^13.2^22-2^26
=3^2.2^36/11.2^35-2^36
=3^2.2^36/2^35-(11-2)
=9.2/9=2
D,
=1152-374-1152+(-65)+374
=(1152-1152)+(-374+374)+(-65)
=0+0+(-65)=-65
E,
=13-(12-11-10+9)+(8-7-6+5)-(4-3-2+1)
=13-0+0-0=13
Bài 2:
A,(19x+50)÷14=25-16
(19x+50)÷14=9
19x+50=126
19x=76
x=4
B,
31x+(1+2+3+...+30)=1240
31x+465=1240
31x=775
×=25
a, =\(3^4+2^5=81+32=113\)
b, =\(3.\left(4^2-2.3\right)=3.\left(16-6\right)=3.10=30\)
c, =\(\dfrac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}=\dfrac{2^{12}.3^{14}}{2^{12}.3^{12}}=3^2=9\)
d, =\(\dfrac{3^2.7^2.2.7.5^3}{5^3.7^3.2.3}=3\)
e, =\(\dfrac{3^6.5^3.2^8.5^4.2^2.3^4}{2^{10}.3^{10}.5^5}=\dfrac{3^{10}.2^{10}.5^7}{2^{10}.3^{10}.5^5}=5^2=25\)
g, =\(\dfrac{2^5.\left(2^8+1\right)}{2^2.\left(2^8+1\right)}=\dfrac{2^5}{2^2}=2^3=8\)
a) 42 . 57 + 43 . 42 - 600
= 2394 + 1806 - 600
= 4200 - 600
= 3600.
b) 22 . 52 - 64 : 23
= 4 . 25 - 64 : 8
= 100 - 8
= 92
c) 120 - [ 100 - ( 5 - 2 )3 ]
= 120 - [ 100 - ( 5 - 2 )3 ]
= 120 - [ 100 - 33 ]
= 120 - [ 100 - 27 ]
= 120 - 73
= 47.
1.Thực hiện các phép tính sau
a)42.57+43.42-600
b)22.52-64: 23
c) 120-[100-(5-2)3]
d)(102+112+122):(132+142)
a) 42 . 57 + 43 . 42 - 600
= 2394 + 1806 - 600
= 4200 - 600
= 3600.
b) 22 . 52 - 64 : 23
= 4 . 25 - 64 : 8
= 100 - 8
= 92
c) 120 - [ 100 - ( 5 - 2 )3 ]
= 120 - [ 100 - ( 5 - 2 )3 ]
= 120 - [ 100 - 33 ]
= 120 - [ 100 - 27 ]
= 120 - 73
= 47.
a, -418-{-218-[-118-(-318)+2012]}
= -418-[-218-(-118+318+2012)]
= -418-(-218+118-318-2012)
= -418+218-118+318+2012
= (218-118)+(318-418)+2012
= 100-100+2012
= 2012
b, 1-2+3-4+...+99-100
Tổng F có số số hạng là:
(100-1):1+1=100(số)
Có số cặp là:
100:2=50(cặp)
Ta có: 1-2+3-4+...+99-100
= (1-2)+(3-4)+...+(99-100)
= (-1)+(-1)+...+(-1)
= (-1).50
=-50
e, 2-5+8-11+14-17+...+98-101
Tổng I có số số hạng là:
(101-2):1+1=100(số)
Có số cặp là:
100:2=50(cặp)
Ta có: 2-5+8-11+14-17+...+98-101
= (2-5)+(8-11)+(14-17)+...+(98-101)
= (-3)+(-3)+(-3)+...+(-3)
= (-3).50
= -150
b/1.2.3.....2013 - 1.2.3....2012-1.2.3...20122
= (1.2.3...2012.2013-1.2.3...2012)-1.2.3...2012.2012
= 1.2.3...2012(2013-1-2012)
=1.2.3...2012.0=0
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
công thức đặc biệt của rubik 5x5 là j
sos
-418-{-218-[-118+318+2024]}
-418-{-218+118-318-2024}
-418+218-118+318+2024
(-418+318)+(218-118)+2024
(-100)+100+2024
2024
a: \(M=-418-\left\{-218-\left[-118-\left(-318\right)+2024\right]\right\}\)
\(=-418-\left\{-218-\left[-118+318+2024\right]\right\}\)
\(=-418-\left\{-218-\left[200+2024\right]\right\}\)
\(=-418-\left\{-218-200-2024\right\}\)
\(=-418+218+200+2024\)
=2024
b: \(2024^2-2023\cdot2024\)
\(=2024\left(2024-2023\right)\)
\(=2024\cdot1=2024\)
c: \(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
\(=\dfrac{100+121+144}{169+196}\)
\(=\dfrac{365}{365}=1\)
d: \(\dfrac{2^6\cdot18+2^7}{2^6\cdot5^2-2^6\cdot3}=\dfrac{2^7\cdot9+2^7}{2^6\left(5^2-3\right)}\)
\(=\dfrac{2^7\left(9+1\right)}{2^6\cdot22}=\dfrac{2\cdot10}{22}=\dfrac{20}{22}=\dfrac{10}{11}\)
B)2024.(2024-2023)
2024.1
2024
=-418-{-218-[-118+318+2024]}
=-418-{-218+118-318-2024}
=-418+218-118+318+2024
A)=(-418+318)+(218-118)+2024
=-100+100+2024
=0+2024
=2025