Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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: \(UCLN\left(2^2\cdot5;2\cdot3\cdot5\right)=2\cdot5=10\)
a ) 22.5 và 2.3.5
ưc : 5 và 2
vậy ƯCLN là 5
b ) 24.3 ; 22.32.5 và 24.11
ưc : 2
ta lấy số 2 có lũy thừa nhỏ nhất là : 22
mà 22= 4
vậy ƯCLN là 4
Bài 1
a) \(x=x^5\)
\(x^5-x=0\)
\(x\left(x^4-1\right)=0\)
\(x=0\) hoặc \(x^4-1=0\)
* \(x^4-1=0\)
\(x^4=1\)
\(x=1\)
Vậy x = 0; x = 1
b) \(x^4=x^2\)
\(x^4-x^2=0\)
\(x^2\left(x^2-1\right)=0\)
\(x^2=0\) hoặc \(x^2-1=0\)
*) \(x^2=0\)
\(x=0\)
*) \(x^2-1=0\)
\(x^2=1\)
\(x=1\)
Vậy \(x=0\); \(x=1\)
c) \(\left(x-1\right)^3=x-1\)
\(\left(x-1\right)^3-\left(x-1\right)=0\)
\(\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)
\(x-1=0\) hoặc \(\left(x-1\right)^2-1=0\)
*) \(x-1=0\)
\(x=1\)
*) \(\left(x-1\right)^2-1=0\)
\(\left(x-1\right)^2=1\)
\(x-1=1\) hoặc \(x-1=-1\)
**) \(x-1=1\)
\(x=2\)
**) \(x-1=-1\)
\(x=0\)
Vậy \(x=0\); \(x=1\); \(x=2\)
Bài 1:
a) \(4^{x+2}+4^x=68\)
\(\Rightarrow4^x\cdot\left(4^2+1\right)=68\)
\(\Rightarrow4^x\cdot17=68\)
\(\Rightarrow4^x=\dfrac{68}{17}\)
\(\Rightarrow4^x=4\)
\(\Rightarrow4^x=4^1\)
\(\Rightarrow x=1\)
b) \(5\cdot2^{x+4}-3\cdot2^x=308\)
\(\Rightarrow2^x\cdot\left(5\cdot2^4-3\right)=308\)
\(\Rightarrow2^x\cdot\left(5\cdot16-3\right)=308\)
\(\Rightarrow2^x\cdot77=308\)
\(\Rightarrow2^x=\dfrac{308}{77}\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\)
\(\Rightarrow x=2\)
c) \(4\cdot3^{x+1}+7\cdot3^x=513\)
\(\Rightarrow3^x\cdot\left(4\cdot3+7\right)=513\)
\(\Rightarrow3^x\cdot19=513\)
\(\Rightarrow3^x=\dfrac{513}{19}\)
\(\Rightarrow3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
d) \(5^{x+4}-5^x=3120\)
\(\Rightarrow5^x\cdot\left(5^4-1\right)=3120\)
\(\Rightarrow5^x\cdot\left(625-1\right)=3120\)
\(\Rightarrow5^x\cdot624=3120\)
\(\Rightarrow5^x\cdot\dfrac{3120}{624}\)
\(\Rightarrow5^x=5\)
\(\Rightarrow5^x=5^1\)
\(\Rightarrow x=1\)
f) \(3\cdot4^{2x+1}-16^x=2816\)
\(\Rightarrow3\cdot4^{2x+1}-\left(4^2\right)^x=2816\)
\(\Rightarrow3\cdot4^{2x+1}-4^{2x}=2816\)
\(\Rightarrow4^{2x}\cdot\left(3\cdot4-1\right)=2816\)
\(\Rightarrow4^{2x}\cdot11=2816\)
\(\Rightarrow4^{2x}=\dfrac{2816}{11}\)
\(\Rightarrow4^{2x}=256\)
\(\Rightarrow\left(2^2\right)^{2x}=2^8\)
\(\Rightarrow2^{4x}=2^8\)
\(\Rightarrow4x=8\)
\(\Rightarrow x=2\)
Bài 2:
\(2^x+124=5^y\)
\(\Rightarrow5^y-2^x=124\)
\(\Rightarrow5^y-2^x=125-1\)
\(\Rightarrow\left\{{}\begin{matrix}5^y=125\\2^x=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5^y=5^3\\2^x=2^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=3\\x=0\end{matrix}\right.\)
Vậy: ....
a: Ta có: \(60-3\left(x-2\right)=51\)
\(\Leftrightarrow3\left(x-2\right)=9\)
\(\Leftrightarrow x-2=3\)
hay x=5
b: Ta có: \(4x-20=2^5:2^3\)
\(\Leftrightarrow4x=24\)
hay x=6
=>\(2^x\cdot16+2^x=5\cdot2^x\)
=>17*2^x-5*2^x=0
=>2^x=0
=>\(x\in\varnothing\)