
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


a) 4x3+15=47
=> 4x3=47-15
=> 4x3=32
=> x3=32:4
=> x3=23
=> x=2
b) 4.2x-3=125
=> 4.2x=125+3
=> 2x=128:4
=> 2x=32
=> 2x=25
=> x=5
4x3 + 15 = 47
4x3 = 47 - 15
4x3 = 32
x3 = 32 : 4
x3 = 8
x3 = 23
x = 2
b) 4 . 2x - 3 = 125
4 . 2x = 125 + 3
4 . 2x = 128
2x = 128 : 4
2x = 32
2x = 25
x = 5

4 . x3 + 15 = 47
4 . x3 = 47 - 15
4 . x3 = 32
x3 = 32 : 4
x3 = 8
=> x = 2
4 . 2x - 3 = 125
4 . 2x = 125 + 3
4 . 2x = 128
2x = 128 : 4
2x = 32
=> x = 5

a)4x3+15=47
4x3=47-15
4x3=32
x3=32:4
x3=8
x3=23
vậy x = 2
b)4.2x-3=125
4.2x=125+3
4.2x=128
2x=128:4
2x=32
2x=25
Vậy x=5
a) 4x3 + 15 = 47 b) 4.2x - 3 = 125
4x3 = 47 - 15 4.2x = 125 + 3
4x3 = 32 4.2x = 128
x3 = 32 : 4 2x = 128 : 4
x3 = 8 2x = 32
x3 = 23 2x = 25
=> x = 2 => x = 5

a, \(4x^2+15=47\\ =>4x^2=47-15=32\\ =>x^2=32:4=8\\ =>x\ne N\)
b, \(4.2^x-3=125\\ =>4.2^x=125+3=128\\ =>2^x=128:4=32\\ Mà:2^5=32\\ =>x=5\)
a) 4\(x^2\)+ 15 = 47
=> 4\(x^2\) = 47 - 15
=> 4\(x^2\)= 32
=> \(x^2\)= 32 : 4
=> \(x^2\)= 8
=> \(x\) \(\ne\) N
b) 4. \(2^x\) - 3 = 125
=> 4.2\(^x\)= 125 + 3
=> 4. \(2^x\)= 128
=> \(2^x\)= 128 : 4
=> \(2^x\)= 32
Mà \(2^5\) = 32 nên \(x\) = 5


a: \(2^{x}\cdot4=128\)
=>\(2^{x}=\frac{128}{4}=32=2^5\)
=>x=5
b: \(x^{15}=x\)
=>\(x^{15}-x=0\)
=>\(x\left(x^{14}-1\right)=0\)
=>\(\left[\begin{array}{l}x=0\\ x^{14}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^{14}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\end{array}\right.\)
c: \(\left(2x+1\right)^3=125\)
=>\(\left(2x+1\right)^3=5^3\)
=>2x+1=5
=>2x=5-1=4
=>\(x=\frac42=2\)
d: \(\left(x-5\right)^4=\left(x-5\right)^6\)
=>\(\left(x-5\right)^6-\left(x-5\right)^4=0\)
=>\(\left(x-5\right)^4\cdot\left\lbrack\left(x-5\right)^2-1\right\rbrack=0\)
=>\(\left(x-5\right)^4\cdot\left(x-5-1\right)\left(x-5+1\right)=0\)
=>\(\left(x-5\right)^4\cdot\left(x-6\right)\left(x-4\right)=0\)
=>\(\left[\begin{array}{l}x-5=0\\ x-6=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=6\\ x=4\end{array}\right.\)

a ) 2 x . 4 = 128
2 x = 32
2 x = 2 5
=> x = 5
b ) x 15 = x
=> x = 0 hoặc x = 1
c ) ( 2x + 1 ) 3 = 125
( 2x + 1 ) 3 = 5 3
=> 2x + 1 = 5
2x = 4
x = 2
d ) ( x - 5 ) 4 = ( x - 5 ) 6
=> x - 5 = 0 hoặc x - 5 = 1
x = 5 x = 6
Vậy x = 5 hoặc x = 6
a) x=5
b) x=1 hoặc x=0 hoặc x=-1
c) x=2
d) x=5 hoặc x=-4 hoặc x=6
\(4\cdot x^3+15=47\) \(4\cdot2^x-3=125\)
\(4x^3=47-15\) \(4\cdot2^x=125+3\)
\(4x^3=32\) \(4\cdot2^x=128\)
\(x^3=32:4\) \(2^x=128:4\)
\(x^3=8\) \(2^x=32\)
\(x^3=2^3\) \(2^x=2^5\)
\(\Rightarrow x=2\) \(\Rightarrow x=5\)
a) 4* x^3 + 15 = 47
4* x^3 =47-15
4* x^3 =32
x^3 =32:4
x^3 =8
Suy ra: 2^3 =8
b)4* 2^x - 3 =125
4* 2^x =125+3
4* 2^x =128
2^x =128:4
2^x =32
Suy ra: 2^4 =32