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1a.4,9,16,25,36,49,64,91,110,121,144,169,400,900,1600
b. 8,27,64,125,1000
C1111^2=1234321
2. x=6 ,x=3,x=0, x k thuộc n, x k thuộc n
a;2 mu x =2 mu 6 .Vậy x=6
- b,7 mu x=49
- 7mu x -1x=7 mu 2
- 7mu x=2-1
- x=1
- c,(x+1)mu 5=2 mu 5
- x+1=2
- x=2+1
- x=3
a) 5x = 125
5x = 53
=> x = 3
b) x3 = 64
x3 = 43
=> x = 4
c) ( x - 1 ) 2 = 25
( x - 1 ) 2 = 52
=> x - 1 = 5
=> x = 5 + 1
=> x = 6
d) 5x + 5x+2 = 130
5x . 1 + 5x . 52 = 130
5x . ( 1 + 52 ) = 130
5x . 26 = 130
5x = 130 : 26
5x = 5
=> x = 1
Ta có: \(4^{2x+4}=4^3\)
\(\Rightarrow\)\(2x+4=3\)
\(\Rightarrow\)\(x=\frac{-1}{2}\)
Ta có: \(3^{x-1}+3^{x-2}=108\)
\(\Rightarrow\)\(3^{x-1}+3^{x-2}=2^2.3^3\)
\(\Rightarrow\)\(3^{x-2}=3^3\)
\(\Rightarrow\)\(x-2=3\)
\(\Rightarrow x=5\)
a: \(3^x-2=2^7\)
\(\Leftrightarrow3^x=128+2=130\)(vô lý)
b: \(4^{x+1}=64\)
=>x+1=3
hay x=2
c: \(\left(5x+1\right)^2=1^{2016}=1\)
=>5x+1=1 hoặc 5x+1=-1
=>x=0 hoặc x=-2/5
d: \(2^{2\left(x-1\right)}=8\)
=>2(x-1)=3
=>x-1=3/2
hay x=5/2
a) \(3^2.x+2^3.x=51\)
\(\Leftrightarrow x\left(3^2+2^3\right)=51\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\)
Vậy
b) \(6^2.2-\left(84-3^2.x\right):7=69\)
\(\Leftrightarrow\left(84-3^2.x\right):7=3\)
\(\Leftrightarrow84-3^2.x=21\)
\(\Leftrightarrow3^2.x=63\)
\(\Leftrightarrow x=7\)
Vậy
Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
a: =>2x^3=58-4=54
=>x^3=27
=>x=3
b; =>(5-x)^5=2^5
=>5-x=2
=>x=3
c: =>(5x-6)^3=4^3
=>5x-6=4
=>5x=10
=>x=2
d: (3x)^3=(2x+1)^3
=>3x=2x+1
=>x=1
1=>2x3=54
=>x3=27 =>x=3
2=>(5-x)5=25
=>5-x=2
=>x=3
3=>(5x-6)3=43
=>5x-6=4
=>5x=10=>x=2
4=>3x=2x+1
=>x=1