(4x-1)^30=(4x-1)^20 Tìm x
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\(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
\(4x^{30}-1^{30}=4x^{20}-1^{20}\)
\(4x^{30}-4x^{20}=-1+1\)
\(4x^{20}\left(x^{10}-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4x^{20}=0\\x^{10}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x^{20}=0\\x^{10}=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}}\)
hok tốt!!
Ta có \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
<=> \(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
<=> \(\left(4x-1\right)^{20}.\left[\left(4x-1\right)^{10}-1\right]=0\)
<=> \(\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=1\\4x-1=1;4x-1=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\4x=2;4x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=2;x=0\end{cases}}\)
Vậy \(x\in\left\{0;2;\frac{1}{4}\right\}\)
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Hc tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
\(\Leftrightarrow\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
\(\Leftrightarrow\left(4x-1\right)^{20}\left[\left(4x-1\right)^{10}-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=1\\4x-1=-1;1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=0\\x=\frac{1}{2}\end{cases}}\)
Vậy \(x=\frac{1}{4};0;\frac{1}{2}\)
P/s : phần \(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=0\\x=\frac{1}{2}\end{cases}}\) thay dấu \(\hept{\begin{cases}\\\\\end{cases}}\) thành dấu \(\orbr{\begin{cases}\\\end{cases}}\) nhé!
\(\Leftrightarrow\orbr{\begin{cases}4x=1\\4x-1=\pm1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}\end{cases}}\)
\(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
\(\Leftrightarrow\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
\(\Leftrightarrow\left(4x-1\right)^{20}.\left[\left(4x-1\right)^{10}-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\4x-1=\pm1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=0\end{cases}}\)
Vậy x = 1/4 hoặc 1/2 hoặc 0
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
Xét \(4x-1=0\)
\(\Rightarrow4x=1\Rightarrow x=\frac{1}{4}\), thỏa mãn
Xét \(4x-1\ne0\)
\(\Rightarrow\left(4x-1\right)^{30}:\left(4x-1\right)^{20}=1\)
\(\Rightarrow\left(4x-1\right)^{10}=1\Rightarrow\left[{}\begin{matrix}4x-1=1\Rightarrow x=\frac{1}{2}\\4x-1=-1\Rightarrow x=0\end{matrix}\right.\)
Vậy....
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :
\(C=\frac{1}{\left|x-2\right|+3}\)
\(C\le\frac{1}{3}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy....
Bài 2 :
a) \(\left(\frac{1}{2}\right)^{3x-1}=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^{3x-1}=\left(\frac{1}{2}\right)^5\)
\(\Rightarrow3x-1=5\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
b) \(2\cdot3^{x-405}=3^{x-1}\)
\(2=3^{x-1}:3^{x-405}\)
\(2=3^{x-1-x+405}\)
\(2=3^{404}\)( vô lí )
=> x thuộc rỗng
c) \(\frac{1}{81}\cdot27^{2x}=\left(-9\right)^4\)
\(\frac{27^{2x}}{81}=9^4\)
\(\frac{\left(3^3\right)^{2x}}{3^4}=\left(3^2\right)^4\)
\(\frac{3^{6x}}{3^4}=3^8\)
\(3^{6x-4}=3^8\)
\(\Rightarrow6x-4=8\)
\(\Rightarrow6x=12\)
\(\Rightarrow x=2\)
d) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
\(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
\(\left(4x-1\right)^{20}\cdot\left[\left(4x-1\right)^{10}-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}4x-1=0\\4x-1=\left\{\pm1\right\}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=\left\{\frac{1}{2};0\right\}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. ta có \(\left(4x+4\right)\left(3y+1\right)=20\Leftrightarrow\left(x+1\right)\left(3y+1\right)=5\)
Vậy \(\hept{\begin{cases}x+1=1\\3y+1=5\end{cases}}\text{ hoặc }\hept{\begin{cases}x+1=5\\3y+1=1\end{cases}}\)
Vì x,y là các số tự nhiên nên ta giải ra \(\hept{\begin{cases}x=4\\y=0\end{cases}}\)
b. ta có 2y+1 là số lẻ và là ước của 30 nên
\(2y+1\in\left\{1,3,5,15\right\}\)
Với \(2y+1=1\Rightarrow x-1=30\Rightarrow\hept{\begin{cases}y=0\\x=31\end{cases}}\)
Với \(2y+1=3\Rightarrow x-1=10\Rightarrow\hept{\begin{cases}y=1\\x=11\end{cases}}\)
Với \(2y+1=5\Rightarrow x-1=6\Rightarrow\hept{\begin{cases}y=2\\x=7\end{cases}}\) Với \(2y+1=15\Rightarrow x-1=2\Rightarrow\hept{\begin{cases}y=7\\x=3\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow124-\left(20-4x\right)=120\\ \Rightarrow20-4x=24\\ \Rightarrow4x=-4\\ \Rightarrow x=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)Ta co: x+20 la boi cua x+2
=>(x+20)chia het cho(x+2)
=>(x+2)+18chia het cho (x+2)
=>18 chia het cho (x+2)
=>(x+2) thuoc Ư(18)
Mà Ư(2)= 1;2;3;6;9;18
ta có bảng sau:
x+2 | 1 | 2 | 3 | 6 | 9 | 18 |
x | ll | 0 | 1 | 4 | 7 | 16 |
Vậy x = 0;1;4;7;16.
Nếu đúng nhớ tặng mình và đúng nhé!
Thank you!
ta có : \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)(=)\(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)(=)\(\left(4x-1\right)^{20}\left[\left(4x-1\right)^{10}-1\right]=0\)(=)\(\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left[\left(4x-1\right)^{10}-1\right]=0\end{cases}}\)(=)\(\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)(=)\(\orbr{\begin{cases}4x=1\\\begin{cases}4x-1=1\\4x-1=-1\end{cases}\end{cases}}\)(=)\(\orbr{\begin{cases}x=\frac{1}{3}\\\begin{cases}4x=2\\4x=0\end{cases}\end{cases}}\)\(\orbr{\begin{cases}x=\frac{1}{4}\\\begin{cases}x=\frac{1}{2}\\x=0\end{cases}\end{cases}}\)