\(\left|2x-3\right|=4x\)
TH1 : \(2x-3>0\Rightarrow\begin{cases}\left|2x-3\right|=2x-3\\x>\frac{3}{2}\end{cases}\)
\(\Rightarrow2x-3=4x\)
\(\Rightarrow2x-4x=3\)
\(\Rightarrow-2x=3\)
\(\Rightarrow x=\frac{-3}{2}\) ( Không thõa mãn )
TH2 : \(2x-3< 0\Rightarrow\begin{cases}\left|2x-3\right|=-\left(2x-3\right)\\x< \frac{3}{2}\end{cases}\)
\(\Rightarrow-\left(2x-3\right)=4x\)
\(\Rightarrow-2x+3=4x\)
\(\Rightarrow-2x-4x=-3\)
\(\Rightarrow-6x=-3\)
\(\Rightarrow x=\frac{1}{2}\left(TM\right)\)
Vậy \(x=\frac{1}{2}\)
|2x-3|=4x
TH1: 2x-3=4x
<=> 2x=-3<=> x=-3/2
TH2: 2x-3=-4x
<=> 6x=3<=> x=1/2
Bài này dùng phương pháp: |A|=B↔A=B hoặc A=-B
Áp dụng:
+TH1: 2x-3=4x→x=-1,5
+TH2: 2x-3=-4x→x=0,5
\(\Rightarrow\left[\begin{array}{nghiempt}2x-3=4x\\-\left(2x-3\right)=4x\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-4x=3\\-2x-4x=-3\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}-2x=3\\-6x=-3\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=\frac{1}{2}\end{array}\right.\)
Vậy \(x=-\frac{3}{2};x=\frac{1}{2}\)
