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\(\left|2x-3\right|-\left|3x+2\right|=0\)
\(\Leftrightarrow\left|2x-3\right|=\left|3x+2\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3x=2+3\\2x+3x=-2+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-x=5\\5x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}\)
Vậy \(x\in\left\{-5;\frac{1}{5}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) TH1: Với \(x< 0\) thì \(\left|2x+3\right|=-\left(2x+3\right)=-2x-3\)
TH2: Với \(x\ge0\) thì \(\left|2x+3\right|=2x+3\)
b) TH1: Với \(x< 0\) thì \(\left|4x-2\right|=-\left(4x-2\right)=-4x+2\)
TH2: Với \(x\ge0\) thì \(\left|4x-2\right|=4x-2\)
c) TH1: Với \(x< 0\) thì \(\left|3x-5\right|=-\left(3x-5\right)=-3x+5\)
TH2: Với \(x\ge0\) thì \(\left|3x-5\right|=3x-5\)
a: TH1: x>=-3/2
=>A=2x+3
TH2: x<-3/2
=>A=-2x-3
b: TH1: x>=1/2
=>A=4x-2
TH2: x<1/2
=>A=-4x+2
c: TH1: x>=5/3
=>B=5x-3
TH2: x<5/3
=>B=-5x+3
Tìm x,y biết
a) 2I2x-3I=\(\frac{1}{2}\)
b)7,5-3I5-2xI=-4,5
c)I3x-4I+I3y+5I=0
d)3,7+I4,3-xI=0
e)4-I5x-2I=1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2\left|2x-3\right|=\frac{1}{2}\)
\(\left|2x-3\right|=\frac{1}{2}:2\)
\(\left|2x-3\right|=\frac{1}{4}\)
\(\orbr{\begin{cases}2x-3=\frac{1}{4}\\2x-3=-\frac{1}{4}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{13}{4}\\2x=\frac{11}{4}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{13}{8}\\x=\frac{11}{8}\end{cases}}\)
b)\(7,5-3\left|5-2x\right|=-4,5\)
\(3\left|5-2x\right|=12\)
\(\left|5-2x\right|=4\)
\(\orbr{\begin{cases}5-2x=4\\5-2x=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=1\\2x=9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{9}{2}\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow\left\{{}\begin{matrix}3x-2>-4\\3x-2< 4\end{matrix}\right.\Leftrightarrow-\dfrac{2}{3}< x< 2\)
c: \(\Leftrightarrow\left[{}\begin{matrix}3x-1>5\\3x-1< -5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -\dfrac{4}{3}\end{matrix}\right.\)
d: \(\Leftrightarrow\left[{}\begin{matrix}3x+1>x-2\\3x+1< -x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x>-3\\4x< 1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{3}{2}\\x< \dfrac{1}{4}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
HS lớp 7 mà ko biết làm bài này người ta nói nó là thằng thiểu năng
a) |2x+1/3|=1/2
\(\Rightarrow\orbr{\begin{cases}2x+\frac{1}{3}=\frac{1}{2}\\2x+\frac{1}{3}=\frac{-1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}2x=\frac{1}{6}\\2x=\frac{-5}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-5}{12}\end{cases}}\)
b) |1-1/2x|=1/3
\(\Rightarrow\orbr{\begin{cases}1-\frac{1}{2}x=\frac{1}{3}\\1-\frac{1}{2}x=\frac{-1}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=\frac{2}{3}\\\frac{1}{2}x=\frac{4}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{8}{3}\end{cases}}\)
c) |3x+1|=1/5
\(\Rightarrow\orbr{\begin{cases}3x+1=\frac{1}{5}\\3x+1=\frac{-1}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}3x=\frac{-4}{5}\\3x=\frac{-6}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-4}{9}\\x=\frac{-2}{5}\end{cases}}\)
d) |x-1/2|+1=5/3
|x-1/2|=5/3-1
|x-1/2|=2/3
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{2}{3}\\x-\frac{1}{2}=\frac{-2}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{6}\\x=\frac{-1}{6}\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét x<-2=>x+2<0=>Ix+2I=-x-2
=>3x+6<3.(-2)+6=>3x+6<0=>I3x+6I=-3x-6
=>I3x+6I-Ix+2I=7
=>-3x-6+x+2=7
=>(-3x+x)-(6-2)=7
=>-2x-4=7
=>-2x=7+4
=>-2x=11
=>x=11:(-2)
=>x=-11/2
Xét x>_-2=>x+2>_0=>Ix+2I=x+2
=>3x+6>_3.(-2)+6=>3x+6>_0=>I3x+6I=3x+6
=>I3x+6I-Ix+2I=7
=>3x+6+x+2=7
=>(3x+x)+(6+2)=7
=>4x+8=7
=>4x=7-8
=>4x=-1
=>x=-1/4
Vậy x=-11/2,-1/4
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ Ta có :
\(E=\left|4x-3\right|+\left|5y+7,5\right|+17,5\)
Mà \(\left\{{}\begin{matrix}\left|4x-3\right|\ge0\\\left|5y+7,5\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow E=\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\)
Để E đạt GTNN thì \(\left\{{}\begin{matrix}\left|4x-3\right|\\\left|5y+7,5\right|\end{matrix}\right.\) đạt GTNN
Dấu "=" xảy ra khi :
\(\left\{{}\begin{matrix}\left|4x-3\right|=0\\\left|5y+7,5\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x-3=0\\5y+7,5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,75\\y=-1,5\end{matrix}\right.\)
Vậy GTNN của E bằng 17,5 khi \(\left\{{}\begin{matrix}x=0,75\\y=-1,5\end{matrix}\right.\)
Ta có: \(|2x-3|-|3x-2|=0\)
\(\Leftrightarrow|2x-3|=|3x-2|\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=3x-2\\2x-3=2-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3x=-2+3\\2x+3x=2+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-x=-1\\5x=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}\Leftrightarrow}x=1\)