Ta có: \(\left|2x-3\right|+\left|\left(2x-3\right)\left(x-1\right)\right|=0\)
\(\Leftrightarrow\left|2x-3\right|+\left|2x-3\right|\cdot\left|x-1\right|=0\)
\(\Leftrightarrow\left|2x-3\right|\cdot\left(\left|x-1\right|+1\right)=0\)
Mà \(\left|x-1\right|+1\ge1>0\left(\forall x\right)\)
\(\Rightarrow\left|2x-3\right|=0\)
\(\Leftrightarrow2x-3=0\)
\(\Rightarrow x=\frac{3}{2}\)
Bài giải
\(\left|2x-3\right|+\left|\left(2x-3\right)\left(x-1\right)\right|=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\\left(2x-3\right)\left(x-1\right)=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=3\\2x-3=0\text{ hoặc }x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{3}{2}\text{ hoặc }x=1\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\frac{3}{2}\text{ ; }1\right\}\)
Vì | 2x - 3 \(\ge\)0\(\forall\)x ; | ( 2x - 3 ) ( x - 1 ) | \(\ge\)0\(\forall\)x
=> | 2x - 3 | + | ( 2x - 3 ) ( x - 1 ) |\(\ge\)0\(\forall\)x
Dấu "=" xảy ra <=> \(\orbr{\begin{cases}\left|2x-3\right|=0\left(1\right)\\\left|\left(2x-3\right)\left(x-1\right)\right|=0\end{cases}}\)
<=> \(\orbr{\begin{cases}2x-3=0\\\left(2x-3\right)\left(x-1\right)=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{3}{2};x=1\end{cases}}\)
Mà x = 1 không làm cho ( 1 ) có giá trị bằng 0
=> x = 3/2 tm đề bài
Bài làm :
Ta có :
\(\hept{\begin{cases}\left|2x-3\right|\ge0\forall x\\\left|\left(2x-3\right).\left(x-1\right)\right|\ge0\forall x\end{cases}}\)
Mà :
|2x-3|+|(2x-3).(x-1)|=0
\(\Rightarrow\hept{\begin{cases}2x-3=0\\\left(2x-3\right)\left(x-1\right)=0\end{cases}}\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy x=3/2
