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1.a) ĐK : \(3-2x\ge0\forall x\Rightarrow x\le\frac{3}{2}\)
Khi đó : \(\left|\frac{1}{2}x\right|=3-2x\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3-2x\\\frac{1}{2}x=-3+2x\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{2}x=3\\\frac{3}{2}x=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=2\end{cases}}\left(tm\right)\)
Vậy \(x\in\left\{\frac{6}{5};2\right\}\)
b) ĐK : \(3x+2\ge0\Rightarrow x\ge\frac{-2}{3}\)
Khi đó : \(\left|x-1\right|=3x+2\Leftrightarrow\orbr{\begin{cases}x-1=3x+2\\x-1=-3x-2\end{cases}}\Rightarrow\orbr{\begin{cases}-2x=3\\4x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1,5\\x=-0,25\left(tm\right)\end{cases}}\)
Vậy x = -0,25
c) ĐKXĐ : \(x-12\ge0\Rightarrow x\ge12\)
Khi đó |5x| = x - 12
<=> \(\orbr{\begin{cases}5x=x-12\\5x=-x+12\end{cases}}\Rightarrow\orbr{\begin{cases}4x=-12\\6x=12\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\left(\text{loại}\right)\)
Vậy \(x\in\varnothing\)
d) ĐK : \(5x+1\ge0\Rightarrow x\ge-\frac{1}{5}\)
Khi đó \(\left|17-x\right|=5x+1\Leftrightarrow\orbr{\begin{cases}17-x=5x+1\\17-x=-5x-1\end{cases}}\Rightarrow\orbr{\begin{cases}6x=16\\-4x=18\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{8}{3}\left(tm\right)\\x=-4,5\left(\text{loại}\right)\end{cases}}\)
Vậy x = 8/3
Tóm lại : Cách làm là
|f(x)| = g(x)
ĐK : g(x) \(\ge0\)
=> \(\orbr{\begin{cases}f\left(x\right)=-g\left(x\right)\\f\left(x\right)=g\left(x\right)\end{cases}}\)
Bạn tự làm tiếp đi ak
![](https://rs.olm.vn/images/avt/0.png?1311)
TH1 : \(x< -4\), ta có:
\(-\left(x+1\right)+\left[-\left(x+4\right)\right]=3x\)
\(-2x-5=3x\)
\(5=-2x-3x\)
\(-5x=5\)
\(\Rightarrow x=-1\) ( Không thỏa mãn \(x< -4\) )
TH2 : \(-4\le x< -1\), ta có:
\(-\left(x+1\right)+\left(x+4\right)=3x\)
\(-x-1+x+4=3x\)
\(3=3x\Rightarrow x=1\)( không thỏa mãn \(-4\le x< -1\))
TH3 : \(x\ge-1\), ta có:
\(x+1+x+4=3x\)
\(2x+5=3x\)
\(\Rightarrow x=5\)(thỏa mãn)
Vậy x=5.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có ; 2|3x - 1| + 1 = 5
<=> 2|3x - 1| = 4
<=> |3x - 1| = 2
<=> \(\orbr{\begin{cases}3x-1=2\\3x-1=-2\end{cases}}\)
a)\(2\left|3x-1\right|+1=5\)
\(TH1:x\ge\frac{3}{5}\).PT có dạng:\(2\left(3x-1\right)+1=5\)
\(\Leftrightarrow6x=6\)
\(\Leftrightarrow x=1\left(TM\right)\)
\(TH2:x< \frac{3}{5}\)Pt có dạng:\(-2\left(3x-1\right)+1=5\)
\(\Leftrightarrow-6x=2\)
\(\Leftrightarrow x=-\frac{2}{6}\left(TM\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(||3x-1|-\dfrac{1}{2}|=\dfrac{5}{2}\)
Có thể xảy ra 2 trường hợp:
TH1:\(||3x-1|-\dfrac{1}{2}|=-\dfrac{5}{2}\)
TH2: \(||3x-1|-\dfrac{1}{2}|=\dfrac{5}{2}\)
Giả sử \(|3x-1|-\dfrac{1}{2}=-\dfrac{5}{2}\)
⇔ \(|3x-1|=-\dfrac{5}{2}+\dfrac{1}{2}\)
⇔ \(|3x-1|=-2\) (Vô lí, vì |3x - 1| ≥ 0 ∀ x)
⇒ \(|3x-1|-\dfrac{1}{2}=\dfrac{5}{2}\)
⇔ \(|3x-1|=\dfrac{5}{2}+\dfrac{1}{2}\)
⇔ \(|3x-1|=3\)
⇔ \(3x-1\in\left\{\pm3\right\}\)
⇔ \(3x\in\left\{-2;4\right\}\)
⇔ \(x\in\left\{-\dfrac{2}{3};\dfrac{4}{3}\right\}\)
Vậy \(x\in\left\{-\dfrac{2}{3};\dfrac{4}{3}\right\}\)
\(\left|\left|3x-1\right|-\dfrac{1}{2}\right|=\dfrac{5}{2}\)
\(\Rightarrow\)2 trường hợp:
Th1:\(3x-1-\dfrac{1}{2}=\dfrac{5}{2}\)
\(3x-1=\dfrac{5}{2}+\dfrac{1}{2}\)
\(3x-1=3\)
\(3x=3+1\)
\(3x=4\Rightarrow x=4:3\Rightarrow x=\dfrac{4}{3}\)
Th2:
\(3x-1-\dfrac{1}{2}=-\dfrac{5}{2}\)
\(3x-1=-\dfrac{5}{2}+\dfrac{1}{2}\)
\(3x-1=-2\)
\(3x=-2+1\)
\(3x=-1\Rightarrow x=-1:3\Rightarrow x=\dfrac{-1}{3}\)
P/s Mình làm theo cách chửa mình nếu sai thì xin lỗi bạn nha