![](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(2x-4)-(x-2)(2x+3)=0`
`<=>2x(x-2)-(x-2)(2x+3)=0`
`<=>(x-2)(2x-2x-3)=0`
`<=>(x-2)*(-3)=0`
`<=>x-2=0`
`<=>x=2`
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left|x-2\right|=x\)
Nếu \(x-2\ge0\Rightarrow x\ge2\)
Thì x - 2 = x => x-x = 2 => 0x=2 => x không có giá trị
Nếu x - 2 < 0 => x < 2
Thì x - 2 = -x => x+x = 2 => 2x = 2 => x = 1 (nhận)
Vậy x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
b
\(\left|6+x\right|\ge0;\left(3+y\right)^2\ge0\Rightarrow\left|6+x\right|+\left(3+y\right)^2\ge0\)
Suy ra \(\left|6+x\right|+\left(3+y\right)^2=0\)\(\Leftrightarrow\hept{\begin{cases}6+x=0\\3+y=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-6\\y=-3\end{cases}}\)
a
Ta có:\(\left|3x-12\right|=3x-12\Leftrightarrow3x-12\ge0\Leftrightarrow3x\ge12\Leftrightarrow x\ge4\)
\(\left|3x-12\right|=12-3x\Leftrightarrow3x-12< 0\Leftrightarrow3x< 12\Leftrightarrow x< 4\)
Với \(x\ge4\) ta có:
\(3x-12+4x=2x-2\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\left(KTMĐK\right)\)
Với \(x< 4\) ta có:
\(12-3x+4x=2x-2\)
\(\Rightarrow10=x\left(KTMĐK\right)\)
![](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)
|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)
c: Ta có: \(\left|\dfrac{1}{2}x-2\right|-\left|x+3\right|=0\)
\(\Leftrightarrow\left|\dfrac{1}{2}x-2\right|=\left|x+3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-2=x+3\\\dfrac{1}{2}x-2=-x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{-1}{2}=5\\x\cdot\dfrac{3}{2}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=-\dfrac{2}{3}\end{matrix}\right.\)