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a <=>12x-4^3=32
<=>12x=96
<=>x=8
b, <=>3x-6=27
<=>3x=33
<=>x=11
c, <=>119-(x-6)=102
<=>x-6=17
<=>x=23
hok tot
a) ( 12x - 43 ) . 83 = 4.84
=> ( 12x - 43 ) : 4 = 84 : 83
=> ( 12x - 43 ) : 4 = 8
=> 12x - 43 = 8 . 4
=> 12x - 64 = 32
=> 12x = 96
=> x = 8
b) ( 3x - 6 ) . 3 = 34
=> ( 3x - 6 ) = 34 : 31
=> 3x - 6 = 33
=> 3x - 6 = 27
=> 3x = 33
=> x = 11
c) 2448 : [ 119 - ( x - 6 ) ] = 24
=> 119 - ( x - 6 ) = 102
=> x - 6 = 7
=> x = 13
Theo đề bài ta có:
(x.5+16):3=7
x.5+16 =7.3
x.5+16 =21
x.5 =21-16
x.5 =5
x =5:5
x =1
Vậy x=1
1/ 3(x - 2) = 9 => x - 2 = 3 => x = 5
2/ 3(x - 36) = 216 => x - 36 = 72 => x = 108
3/ (x - 3)3 = 33 => x - 3 = 3 => x = 6
4/ 2x = 32 => 2x = 25 => x = 5
5/ 3x - 16 = 2.74 : 73 = 14 => 3x = 30 => x = 10
6/ (6x - 39) : 3 = 201 => 6x - 39 = 603 => 6x = 642 => x = 107
7/ (3x - 15)2 = 81 => (3x - 15)2 = 92 => 3x - 15 = 9 => 3x = 24 => x = 8
8/ 3(x + 4) = 105 => x + 4 = 35 => x = 31
9/ 12x - 43 = 4.84 : 83 = 32 => 12x = 32 : 64 => 12x = 1/2 => x = 1/24
10/ 41 - (2x - 5) = 680 => 2x - 5 = -639 => 2x = -634 => x = - 317
a) 27^16 : 9^10
Ta có: (3.9)^16 : 9^10
= 3^16.9^16: 9^10
= 3^16. 9^6
= 3^16.(3^2)^6
=3^16.3^12
=3^28
Tìm x\(\in\)n
a,x3-23=25-(316:314+28:216)
b,5x-2-32=24-(68:66-62)
c,(x2-1)4=81
d,3x+42=196:(193.192)-3.1
a)120-3(x+4)=23
=>3(x+4)=120-23
=>3(x+4)=97
=>x+4=97:3
=>x+4=\(\frac{97}{3}\)
=>x=\(\frac{97}{3}\)-4
=>x=\(\frac{85}{3}\)
Vậy x=\(\frac{85}{3}\)
b)[(4x+28).3+55]:5=35
=>(4x+28).3+55=35.5
=>(4x+28).3+55=175
=>(4x+28).3=175-55
=>(4x+28).3=120
=>4x+28=120:3
=>4x+28=40
=>4x=40-28
=>4x=12
=>x=12:4
=>x=3
Vậyx= 3
c)\((12x-4^3).8^3=4.8^4\)
=>\(12x-64=4.8^4:8^3\)
=>12x-64=4.8
=>12x-64=32
=>12x=96
=>x=96:12
=>x=8
Vậy x=8
d)720:[41-(2x-5)]=23.5
=>720:[41-2x-5)]=40
=>41-(2x-5)=720:40
=>41-(2x-5)=18
=>2x-5=41-18
=>2x-5=23
=>2x=23+5
=>2x=28
=>x=28:;2
=>x=14
Vậy x=14
a, 120 - 3 [ x+4]=23
3 [x+4]= 120-23 = 97
3x = 97-4
3x= 93
x= 93:3
x=31
em bt làm mỗi bài này thui mà ko bt đúng ko
\(\left(12x-4\right)^3.8^3=4^{16}.16^3\)
\(\Leftrightarrow8\left(12x-4\right)^3=4^6.\left(4^2\right)^{^3}\)
\(\Leftrightarrow\left(96x-32\right)^3=4^6.4^6\)
\(\Leftrightarrow\left(96x-32\right)^3=16^6\)
\(\Leftrightarrow\left(96x-32\right)^3=\left(16^2\right)^{^3}\)
\(\Leftrightarrow\left(96x-32\right)^3=256^3\)
\(\Leftrightarrow96x-32=256\)
\(\Leftrightarrow x=\frac{256+32}{96}\)
Vậy \(x=3\)
(12x-4)3.83=43.163
<=> [(12x-4).8]3=(4.16)3
=> (12x-4).8=4.16
<=> 12x-4=8
<=> 12x=12
=> x=1