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3.2x—3=45
3.2x=45+3
3.2x=48
2x=48:3
2x=16
x=16:2
x=8
4x .3+12=120
4x.3=120–12
4x.3=108
4x=108:3
4x=36
x=36:4
x=9
\(3.2x-3=45\)
\(3.2x=45+3\)
\(3.2x=48\)
\(2x=\frac{48}{3}\)
\(2x=16\)
\(x=8\)
Vậy \(x=8\)
\(1500:\left[\left(36x+40\right):x\right]=30\)
\(\left(36x+40\right):x=1500:30\)
\(36x+40=50x\)
\(50x-36x=40\)
\(14x=40\)
\(x=\frac{40}{14}\)
\(x=\frac{20}{7}\)
Vậy \(x=\frac{20}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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a: Ta có: \(\left[\left(6x-39\right):3\right]\cdot28=5628\)
\(\Leftrightarrow\left(6x-39\right):3=201\)
\(\Leftrightarrow6x-39=603\)
\(\Leftrightarrow6x=642\)
hay x=107
b: Ta có: \(4x^3+12=120\)
\(\Leftrightarrow4x^3=108\)
\(\Leftrightarrow x^3=27\)
hay x=3
c: Ta có: \(1500:\left[\left(30x+40\right):x\right]=30\)
\(\Leftrightarrow\left(30x+40\right):x=50\)
\(\Leftrightarrow30x+40=50x\)
hay x=-2
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\(a,223-6\left(x+5\right)=49\\ 6\left(x+5\right)=223-49\\ 6\left(x+5\right)=174\\ x+5=174:6\\ x+5=29\\ x=29-5\\ x=24\\ ----\\ b,4x^3+12=120\\ 4x^3=120-12\\ 4x^3=108\\ x^3=108:4\\ x^3=27\\ Mà:3^3=27\\ Nên:x^3=3^3\\ Vậy:x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(70-\left(x-3\right)=45\)
\(x-3=70-45\)
\(x-3=25\)
\(x=25+3\)
\(x=28\)
b) \(12+\left(5+x\right)=20\)
\(5+x=20-12\)
\(5+x=8\)
\(x=8-5\)
\(x=3\)
c) \(130-\left(100+x\right)=25\)
\(100+x=130-25\)
\(100+x=105\)
\(x=105-100\)
\(x=5\)
d) \(175+\left(30-x\right)=200\)
\(30-x=200-175\)
\(30-x=25\)
\(x=30-25\)
\(x=5\)
e) \(\left(x+12\right)+22=92\)
\(x+12=92-22\)
\(x+12=70\)
\(x=70-12\)
\(x=58\)
f) \(95-\left(x+2\right)=45\)
\(x+2=95-45\)
\(x+2=50\)
\(x=50-2\)
\(x=48\)
a)
70 - (x - 3) = 45
x - 3 = 70 - 45 = 25
x = 25 + 3 = 28
Vậy x = 28
b)
12 + (5 + x) = 20
5 + x = 20 - 12 = 8
x = 8 - 5 = 3
Vậy x = 3
c)
130 - (100 + x) = 25
100 + x = 130 - 25 = 115
x = 115 - 100 = 15
Vậy x = 15
d)
175 + (30 - x) = 200
30 - x = 200 - 175 = 25
x = 30 - 25 = 5
Vậy x = 5
e)
(x + 12) + 22 = 92
x + 12 = 92 - 22 = 70
x = 70 - 12 = 58
Vậy x = 58
f)
95 - (x + 2) = 45
x + 2 = 95 - 45 = 50
x = 50 - 2 = 48
Vậy x = 48
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4^0\cdot9^3-6^9\cdot\dfrac{120}{8^4\cdot3^{12}+6^{11}}\) hay \(4^0\cdot9^3-6^9\cdot\dfrac{120}{8^4}\cdot3^{12}+6^{11}\)?
Tìm x, biết:
a) 4x3+12=120
=>4x3=120-12
=>4x3=108
=>x3=108:4
=>x3=27
=>x3=33
=>x=3
b) 3.2x-3=45
=>3.(2x-1)=45
=>2x-1=45:3
=>2x-1=15
=>2x=15+1
=>2x=16
=>2x=24
=>x=4
4x^3+12=120
4x^3 =120-12
4x^3 =108
x^3 =108:4
x^3 =27
x^3 =3^3
\(\Rightarrow\)x=3
3.2^x-3=45
3.2^x=45+3
3.2^x=48
2^x=48:3
2^x=16
2^x=2^4
\(\Rightarrow\)x=4
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