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\(1,2x+3x-4x=\left(-2\right)^3\)
<=>\(x=-8\)
\(2,x-2x=4^2+4^0\)
<=>\(-x=16+1\)
<=>\(-x=17\)
<=>\(x=-17\)
\(3,2^3x-3^2x=|12-21|\)
<=>\(-x=9\)
<=>\(x=-9\)
\(4,x-45=2x+54\)
<=>\(x-2x=54+45\)
<=>\(-x=99\)
<=>\(x=-99\)
\(5,5x-12+23=6^7:6^5\)
<=>\(5x+11=6^2\)
<=>\(5x+11=36\)
<=>\(5x=25\)
<=>\(x=5\)

a, 95-13(2x-1)=30
<=> 13(2x-1)=65
<=> 2x-1=5
<=> 2x=6
<=> x=3
Vậy..
b, x^2 -4 =21
<=> x^2=25
<=> x^2=5^2
=> x=5 hoặc x=-5
Vậy...
c, 136- (3x-1)^3=12^2
<=> (3x-1)^3=-8
<=> (3x-1)^3=(-2)^3
=> 3x-1=-2
<=> 3x=-1
<=> x= -1/3
Vậy...
a) 95 - 13(2x - 1) = 30
13(2x - 1) = 95 - 30
13(2x - 1) = 65
2x - 1 = 65 : 13
2x - 1 = 5
2x = 5 + 1
2x = 6
x = 6 : 2
x = 3
b) x2 - 4 = 21
x2 = 21 + 4
x2 = 25
Vì 52 = 25 => x = 5
c) 136 - (3x - 1)3 = 122
3x - 1 = 122 - 136
3x - 1 = 8
3x = 8 : 1
3x = 8
x = 8 : 3
x = \(\frac{8}{3}\)

a) = 53+4+1 = 58
b) = 65-4 = 6
c) = 32+3+4-7 = 32
d) = x4+2+1 = x7
e) = 812+10+3 = 825
1. \(5^3.5^4.5=5^{3+4+1}=5^8\)
2. \(6^5:\left(6^3.6\right)=6^5:\left(6^{3+1}\right)=6^5:6^4=6^{5-4}=6^1\)
3. \(\left(3^2.3^3.3^4\right):3^7=\left(3^{2+3+4}\right):3^7=3^9:3^7=3^{9-7}=3^2\)
4. \(x^4.x^2.x=x^{4+2+1}=x^7\)
5. \(\left(8^{12}.8^{10}\right).8^3=8^{12+10+3}=8^{25}\)

56 x 52 = 58
1012 x 102 = 1014
23 x 25 = 28
36 x 32 = 38
41 x 46 = 47
67 x 61 = 68
\(5^6\cdot5^2=5^{6+2}=5^8\)
\(10^{12}\cdot10^2=10^{12+2}=10^{14}\)
\(2^3\cdot2^5=2^{3+5}=2^8\)
\(3^6\cdot3^2=3^{6+3}=3^9\)
\(4^1\cdot4^6=4^{1+6}=4^7\)
\(6^7\cdot6^1=6^{1+7}=6^8\)

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

a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6

a,71.2-6.(2x+5)=10^5:10^3
142-6.(2x+5)=10^2
142-6.(2x+5)=100
6.(2x+5)=142-100
6.(2x+5)=42
2x+5=42:6
2x+5=7
2x=7-5
2x=2
x=1
Vậy x=1

a/ (ghi lại cái đề)
=>+ 3x-7=2
3x=2+7=9
x=3
+ 3x-7=-2
3x=-2+7=5
x=\(\frac{5}{3}\)
b/ (5x-10)2=100
=> +5x-10=10
5x=10+10=20
x=4
+ 5x-10=-10
5x=-10+10=0
x=0
136 - 2x-1=36 : 34 . 4 + 12
136 -2x-1= 32 . 4 + 12
136 - 2x-1= 9 . 4 + 12
136 - 2x-1 = 36 + 12
136 - 2x-1 = 48
2x-1= 136 - 48
2x-1= 88
x - 1 = 88 : 2
x - 1 = 44
x = 44 + 1
x = 45
Vậy x = 45