Tìm số tự nhiên x , biết:
a) 4x^3 + 15 = 47
b)4.2^x - 3 = 125
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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
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
\(4x^3+15=47\)
\(\Leftrightarrow4x^3=32\)
\(\Leftrightarrow x^3=8\)
\(\Leftrightarrow x^3=2^3\)
\(\Leftrightarrow x=2\)
\(4.2^x-3=125\)
\(\Leftrightarrow4.2^x=128\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
a) \(128-3\left(x+4\right)=23\)
\(\Rightarrow3\left(x+4\right)=128-23\)
\(\Rightarrow3\left(x+4\right)=105\)
\(\Rightarrow x+4=35\)
\(\Rightarrow x=35-4\)
\(\Rightarrow x=31\)
b) \(\left[\left(4x+28\right)\cdot3+55\right]:5=35\)
\(\Rightarrow\left(4x+28\right)\cdot3+55=35\cdot5\)
\(\Rightarrow\left(4x+28\right)\cdot3+55=175\)
\(\Rightarrow\left(4x+28\right)\cdot3=120\)
\(\Rightarrow4x+28=40\)
\(\Rightarrow4x=12\)
\(\Rightarrow x=3\)
a, \(128-3\left(x+4\right)=23\)
\(=>3\left(x+4\right)=128-23\)
\(=>3\left(x+4\right)=105\)
\(=>x+4=105:3\)
\(=>x+4=35\)
\(=>x=35-4\)
\(=>x=31\)
b, \(\left[\left(4x+28\right).3+55\right]:5=35\)
\(=>\left(4x+28\right).3+55=35.5\)
\(=>\left(4x+28\right).3+55=175\)
\(=>\left(4x+28\right).3=175-55\)
\(=>\left(4x+28\right).3=120\)
\(=>4x+28=120:3\)
\(=>4x+28=40\)
\(=>4x=40-28\)
\(=>4x=12\)
\(=>x=12:4\)
\(=>x=3\)
\(#WendyDang\)
a) \(4x^3+15=47\)
\(\Rightarrow4x^3=32\)
\(\Rightarrow x^3=8\)
\(\Rightarrow x^3=2^3\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
b) \(4.2^x-3=125\)
\(\Rightarrow4.2^x=128\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
a ) \(4x^3+15=47\)
\(\Leftrightarrow4x^3=32\)
\(\Leftrightarrow x^3=8\)
\(\Leftrightarrow x^3=2^3\)
\(\Leftrightarrow x=3\)
\(4.2^x-3=125\)
\(\Leftrightarrow4.2^x=128\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)