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1. ( 5 ; 2 )
2. a = 364
b = 81
c = 3
Đúng thì tk cho mk nha !!!
2x-1 + 4 . 2x = 288
=> 2x = 288 : 4
2x = 72
2x = 26
x = 6
x3 = 1252
x3 = 15625
x3 = 253
x = 25
1020 < a < 1021
100000000000000000000 < a < 1000000000000000000000
Vì 1020 có 21 chữ số, 1021 có 22 chữ số nên a có 21 chữ số
Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
Lời giải:
a. $x^3=4^3\Rightarrow x=4$
b. $x^2=49=7^2=(-7)^2$
$\Rightarrow x=7$ hoặc $x=-7$
c. $x^3+1=28$
$x^3=28-1=27=3^3$
$\Rightarrow x=3$
d. $2^x=16=2^4$
$\Rightarrow x=4$
e. $2^4.2^x=2^6$
$\Rightarrow 2^{4+x}=2^6$
$\Rightarrow 4+x=6$
$\Rightarrow x=2$
g.
$5^x=25.5^3=5^2.5^3=5^5$
$\Rightarrow x=5$
Lần sau bạn lưu ý viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để đề được rõ ràng hơn nhé.
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
Bo may la binh day k di hieu ashdbfgbgygygggydfsghuyfhdguuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu3
`(2^x+1)^2 =25`
`=> (2^x+1)^2 = (+-5)^2`
\(\Rightarrow\left[{}\begin{matrix}2^x+1=5\\2^x+1=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2^x=4\\2^x=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x\in\varnothing\end{matrix}\right.\)
\(\left(x+6\right)\left(5^x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+6=0\\5^x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\5^x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=0\end{matrix}\right.\)
\(\left(x-3\right)^{2023}=x-3\)
\(\Rightarrow\left(x-3\right)^{2023}-\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left[\left(x-3\right)^{2022}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{2022}-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\\left(x-3\right)^{2022}=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
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 \(x=0\); \(x=1\); \(x=2\)
\(3.2^x+2^{x+1}+2^{x+2}=288\\ \Leftrightarrow2^x\left(3+2+2^2\right)=288\\ \Leftrightarrow2^x=32=2^5\\ \Leftrightarrow x=5\)
tik mik nha
Ta có: \(3\cdot2^x+2^{x+1}+2^{x+2}=288\)
\(\Leftrightarrow2^x\cdot\left(3+2+4\right)=288\)
\(\Leftrightarrow2^x=32\)
hay x=5