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Bài 9,
62x73+36x33=36x73+36x27=36(73+27)=36x100=3600.
197-\([\)6x(5-1)2+20220\(]\):5=197-\([\)6x16+1\(]\):5=197-97:5=197-97/5=888/5.
Bài 10,
21-4x=13
=>4x=21-13=8
=>x=8:4=2.
30:(x-3)+1=45:43=42=16
=>30:(x-3)=16-1=15
=>x-3=30:15=2
=>x=2+3=5.
(x-1)3+5x6=38
=>(x-1)3+30=38
=>(x-1)3=38-30=8=23
=>x-1=2
=>x=3.
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
a, \(390-\left(x-7\right)=13^2:12\)
\(390-\left(x-7\right)=\) \(\dfrac{169}{12}\)
\(x-7=390-\dfrac{169}{12}\)
\(x-7=\dfrac{4511}{12}\)
\(x=\dfrac{4511}{12}+7\)
\(x=\dfrac{4595}{12}\)
Vậy ...
b, \(\left(x-35.2^2\right):7=3^3-24\)
\(\left(x-35.4\right):7=27-24\)
\(\left(x-140\right):7=3\)
\(\Leftrightarrow\left(x-140\right)=3.7\)
\(\Leftrightarrow x-140=21\)
\(\Leftrightarrow x=161\)
Vậy .....
c) \(x-6:2-\left(4^2.3-24\right):2:6=3\)
\(x-3-\left(16.3-24\right):2:6=3\)
\(x-3-\left(48-24\right):2:6=3\)
\(x-3-24:2:6=3\)
\(x-3-2=3\)
\(x=3+2+3\)
\(x=8\)
Vậy ......
d) \(4x-5=5+5^2+5^3+.....+5^{99}\)
Đặt :
\(A=5+5^2+.........+5^{99}\)
\(\Leftrightarrow5A=5^2+5^3+..........+5^{100}\)
\(\Leftrightarrow5A-A=\left(5^2+5^3+......+5^{100}\right)-\left(5+5^2+....+5^{99}\right)\)
\(\Leftrightarrow4A=5^{100}-5\)
\(\Leftrightarrow A=\dfrac{5^{100}-5}{4}\)
\(\Leftrightarrow4x+5=\dfrac{5^{100}-5}{4}\)
Đến đây thì sao nữa nhỉ ?
e) \(\left(2x-1\right)^4=625\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=5\\\left(2x-1\right)^4=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy ....
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 \(x=0\); \(x=1\); \(x=2\)