![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(x^{23}=64.x^{20}\)
\(\Leftrightarrow\frac{x^{23}}{x^{20}}=64\)
\(\Leftrightarrow x^3=64\Rightarrow x=4\)
b)\(\left(4x-3\right)^4=3-4x\)
\(\Leftrightarrow\left(3-4x\right)^4=3-4x\)
\(\Leftrightarrow\left(3-4x\right)^3=1\)
\(\Leftrightarrow3-4x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`#3107.101107`
`|3x - 1| = x + 2`
`\Rightarrow` TH1: `3x - 1 = x + 2`
`\Rightarrow 3x - x = 2 + 1`
`\Rightarrow 2x = 3`
`\Rightarrow x =` $\dfrac{3}2$
TH2: `3x - 1 = -(x + 2)`
`\Rightarrow 3x - 1 = -x - 2`
`\Rightarrow 3x + x = -2 + 1`
`\Rightarrow 4x = -1`
`\Rightarrow x =` $\dfrac{-1}4$
Vậy, \(x\in\left\{-\dfrac{1}{4};\dfrac{3}{2}\right\}.\)
|3x - 1| = x + 2
*) TH1: x ≥ 1/3, ta có:
|3x - 1| = x + 2
3x + 1 = x + 2
3x - x = 2 - 1
2x = 1
x = 1/2 (nhận)
*) TH2: x < 1/3, ta có:
|3x - 1| = x + 1
1 - 3x = x + 1
-3x - x = 1 - 1
-4x = 0
x = 0 (nhận)
Vậy x = 0; x = 1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{1}{2}-\dfrac{x}{3}\right)^2=\dfrac{36}{49}\\ \Rightarrow\left(\dfrac{1}{2}-\dfrac{x}{3}\right)^2=\left(\dfrac{6}{7}\right)^2\\ \Rightarrow\dfrac{1}{2}-\dfrac{x}{3}=\pm\dfrac{6}{7}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-\dfrac{x}{3}=\dfrac{6}{7}\\\dfrac{1}{2}-\dfrac{x}{3}=-\dfrac{6}{7}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{5}{14}\\\dfrac{x}{3}=\dfrac{19}{14}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{14}\times3\\x=\dfrac{19}{14}\times3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{15}{14}\\x=\dfrac{57}{14}\end{matrix}\right.\)
\(\left(3-\dfrac{2}{3}x\right)^3=-\dfrac{1}{64}\\ \Rightarrow\left(3-\dfrac{2}{3}x\right)^3=\left(-\dfrac{1}{4}\right)^3\\ \Rightarrow3-\dfrac{2}{3}x=-\dfrac{1}{4}\\ \Rightarrow\dfrac{2}{3}x=3-\left(-\dfrac{1}{4}\right)\\ \Rightarrow\dfrac{2}{3}x=\dfrac{13}{4}\\ \Rightarrow x=\dfrac{13}{4}:\dfrac{2}{3}\\ \Rightarrow x=\dfrac{13}{4}\times\dfrac{3}{2}\\ \Rightarrow x=\dfrac{39}{8}\)
Hic 2 câu em làm dr xong tự nhiên thử lung tung rồi lại xóa bài ;-;
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-1\right)^2+\left(y+2\right)^2=0\)
Vì \(\left(x-1\right)^2\ge0\forall x\)
\(\left(y+2\right)^2\ge0\forall y\)
Nên \(\left(x-1\right)^2+\left(y+2\right)^2\ge0\) \(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-1=0\\y+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\y=-2\end{cases}}\)
Vậy x = 1 và y = -2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x-\frac{1}{2}\right)\left(2x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=0\\2x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)
b) \(\left(x-\frac{1}{2}\right)\left(x+2\right)< 0\)
TH1: \(\hept{\begin{cases}x-\frac{1}{2}< 0\\x+2>0\end{cases}\Rightarrow\hept{\begin{cases}x< \frac{1}{2}\\x< -2\end{cases}}}\)
TH2: \(\hept{\begin{cases}x-\frac{1}{2}>0\\x+2< 0\end{cases}\Rightarrow\hept{\begin{cases}x>\frac{1}{2}\\x< -2\end{cases}}}\)
x=6 nha!
\(\left(\frac{1}{2}\right)^x=\frac{1}{64}\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^6\)
=> x=6