a.
2x . 4 = 128
2x = 128 : 4
2x = 32
2x = 25
x = 5
b.
x15 = x
Vậy x = 0 hoặc x = 1 hoặc x = -1
c.
(2x + 1)3 = 125
(2x + 1)3 = 53
2x + 1 = 5
2x = 5 - 1
2x = 4
x = 4 : 2
x = 2
d.
(x - 5)4 = (x - 5)6
TH1:
x - 5 = 0
x = 5
TH2:
x - 5 = -1
x = -1 + 5
x = 4
TH2:
x - 5 = 1
x = 1 + 5
x = 6
Vậy x = 5 hoặc x = 4 hoặc x = 6
Chúc bạn học tốt ^^
a) \(2^x.4=128\)
=> \(2^x=32\) => \(2^x=2^5\) => x = 5
b) \(x^{15}=x\) => x = 1 hoặc x = 0
c) \(\left(2x+1\right)^3=125\)
=> \(\left(2x+1\right)^3=5^3\) => 2x + 1 = 5 => x = 2
d) \(\left(x-5\right)^4=\left(x-5\right)^6\)
=> x - 5 = 0 hoặc x - 5 = 1
=> x = 5 hoặc x= 6
Chúc bạn làm bài tốt
\(a,2^x.4=128\)
\(2^x\) \(=128:4\)
\(2^x\) \(=32\)
\(2^x\) \(=2^5\)
\(x\) \(=5\)
\(b,x^{15}=x\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)
\(c,(2x+1)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1\) \(=5\)
\(2x\) \(=5-1\)
\(2x\) \(=4\)
\(x\) \(=4:2\)
\(x\) \(=2\)
\(d,\left(x-5\right)^4=(x-5)^6\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=-1\\x-5=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0+5\\x=\left(-1\right)+5\\x=1+5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
