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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 \(x=0\); \(x=1\); \(x=2\)
a) \(4^n=4096\Rightarrow4^n=4^6\Rightarrow n=6\)
b) \(5^n=15625\Rightarrow5^n=5^6\Rightarrow n=6\)
c) \(6^{n+3}=216\Rightarrow6^{n+3}=6^3\Rightarrow n+3=3\Rightarrow n=0\)
d) \(x^2=x^3\Rightarrow x^3-x^2=0\Rightarrow x^2\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
e) \(3^{x-1}=27\Rightarrow3^{x-1}=3^3\Rightarrow x-1=3\Rightarrow x=4\)
f) \(3^{x+1}=9\Rightarrow3^{x+1}=3^2\Rightarrow x+1=2\Rightarrow x=1\)
g) \(6^{x+1}=36\Rightarrow6^{x+1}=6^2\Rightarrow x+1=2\Rightarrow x=1\)
h) \(3^{2x+1}=27\Rightarrow3^{2x+1}=3^3\Rightarrow2x+1=3\Rightarrow2x=2\Rightarrow x=1\)
i) \(x^{50}=x\Rightarrow x^{50}-x=0\Rightarrow x\left(x^{49}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}=1=1^{49}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
4n = 4096
4n = 212
n = 12
5n = 15625
5n = 56
n = 6
6n+3 = 216
6n+3 = 23.33
6n+3 = 63
n + 3 = 3
Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
1) \(2^x=4^3 \Leftrightarrow2^x=2^6\Leftrightarrow x=6\)
2) \(2^x=4^6\Leftrightarrow2^x=2^{12}\Leftrightarrow x=12\)
3) \(3^x=9^{10}\Leftrightarrow3^x=3^{20}\Leftrightarrow x=20\)
\(\text{a) ( x + 1 ) + ( x + 3 ) + ..... + ( x + 99 ) = 0}\)
\(\Rightarrow\left(x+x+x+.....+x\right)+\left(1+3+5+....+99\right)=0\)
\(\text{Ta có :}\)
\(1+3+5+...+99=\frac{\left(99-1\right):2+1.\left(99+1\right)}{2}=2500\)
\(\Rightarrow50x+2500=0\)
\(\Rightarrow50x=-2500\)
\(\Rightarrow x=-50\)
a. 3x+3x+1+3x+2=1053
=> 3x.(1+3+32)=1053
=> 3x.(1+3+9)=1053
=> 3x.13=1053
=> 3x=1053:13
=> 3x=81
=> 3x=34
=> x=4
b. 5x.519=520.511
=> 519+x=520+11
=> 19+x=20+11
=> 19+x=31
=> x=31-19
=> x=12
\(3^x+3^{x+1}+3^{x+2}=1053\)
\(3^x\left(1+3+3^2\right)=1053\)
\(3^x\cdot13=1053\)
\(3^x=1053:13=81\)
\(3^x=3^4\)
=> x = 4
10.72601467
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