![](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)
`(x-1/2)^2 +(2x-1)^2=0`
\(\Rightarrow\left(x^2-x+\dfrac{1}{4}\right)+\left(4x^2-4x+1\right)=0\\ \Rightarrow x^2-x+\dfrac{1}{4}+4x^2-4x+1=0\\ \Rightarrow5x^2-5x+\dfrac{5}{4}=0\\ \Rightarrow\dfrac{5}{4}\left(4x^2-4x+1\right)=0\\ \Rightarrow\dfrac{5}{4}\left(2x-1\right)^2=0\\ \Rightarrow\left(2x-1\right)^2=0\\ \Rightarrow2x-1=0\\ \Rightarrow2x=0+1\\ \Rightarrow2x=1\\ \Rightarrow x=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x+1\right)\left|x-3\right|=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\\left|x-3\right|=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(\left(x-\frac{1}{2}\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=0\\2x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
|x - 4| + |6 - x| = 0
|x - 4| ; |6 - x| \(\ge\) 0
=> |x - 4| = |6 - x| = 0
|x - 4| = 0 => x= 4
|6 - x| = 0 => x= 6
Vì \(4\ne6\) n ê n không có giá trị của x
Bạn làm các câu khác tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
PT `<=> |x-2|=|1-2x|`
`<=> (x-2)^2=(1-2x)^2`
`<=>x^2-4x+4=1-4x+4x^2`
`<=>x^2+4=1+4x^2`
`<=>3x^2-3=0`
`<=>x= \pm 1`
Vậy `x=-1;x=1`.
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Rightarrow}\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
b. \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=-1\left(Voly\right)\\x=4\end{cases}\Rightarrow x=4}\)
c, \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
d, \(\left(\frac{4}{5}\right)^{5x}=\left(\frac{4}{5}\right)^7\)
\(\Rightarrow5x=7\)
\(\Rightarrow x=\frac{7}{5}\)
e, Ta có: \(A=\frac{x+5}{x-2}=\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)
Để A ∈ Z <=> (x - 2) ∈ Ư(7) = { ±1; ±7 }
x - 2 | 1 | -1 | 7 | -7 |
x | 3 | 1 | 9 | -5 |
Vậy....
a) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
Vậy : ....
b) \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\left(loại\right)\\x=4\end{cases}}\)
c) \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
Vậy :...
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm x biết :a) ( 2x - 3 ).( x +1 ) > 0b) ( x + 5 ).(x-7) < 0c) | 2x - 3 | + 8 = 10d) ( 2x + 5 ) . | x -8 | . ( x2 + 1 ) = 0
sai de
sai de