\(\left|\frac{3}{2}x-1\right|-\left|x+\frac{3}{2}\right|=0\)
Ta có :
\(\left|\frac{3}{2}x-1\right|-\left|x+\frac{3}{2}\right|\ge\left|\frac{3x}{2}-1-x-\frac{3}{2}\right|=\left|\frac{x-5}{2}\right|\)
Tự làm nốt
Ta có: \(\left|\frac{3}{2}x-1\right|-\left|x+\frac{3}{2}\right|=0\)
\(\Leftrightarrow\left|\frac{3}{2}x-1\right|=\left|x+\frac{3}{2}\right|\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}x-1=x+\frac{3}{2}\\\frac{3}{2}x-1=-x-\frac{3}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=\frac{5}{2}\\\frac{5}{2}x=-\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-\frac{1}{5}\end{cases}}\)
\(\left|\frac{3}{2}x-1\right|-\left|x+\frac{3}{2}\right|=0=>\orbr{\begin{cases}\left|\frac{3}{2}x-1\right|=0\\\left|x+\frac{3}{2}\right|=0\end{cases}}\)
\(\orbr{\begin{cases}\frac{3}{2}x=0+1\\x=0-\frac{3}{2}\end{cases}}=>\orbr{\begin{cases}\frac{3}{2}x=1\\x=-\frac{3}{2}\end{cases}}\)
\(\orbr{\begin{cases}x=1:\frac{3}{2}\\x=-\frac{3}{2}\end{cases}}=>\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{3}{2}\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{3}{2}\end{cases}}\)
Đây mà là toán lớp 2 á
đây là toán lớp 2 thì chịu r +_+
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