Tìm x biết:
a) \(\left|\dfrac{1}{2}x\right|=3-2x\)
b) \(\left|x-1\right|=3x+2\)
c) \(\left|5x\right|=x+12\)
a) \(\left|\dfrac{1}{2}x\right|=3-2x\)
+) TH1: \(\dfrac{1}{2}x\ge0\Rightarrow x\ge0\)
Khi đó: \(\dfrac{1}{2}x=3-2x\)
\(\Rightarrow x=2\left(3-2x\right)\)
\(\Rightarrow x=6-4x\)
\(\Rightarrow x+4x=6\)
\(\Rightarrow5x=6\Rightarrow x=\dfrac{6}{5}\) (thỏa mãn)
+) TH2: \(-\dfrac{1}{2}x< 0\Rightarrow x>0\)
\(-\dfrac{1}{2}x=3-2x\)
\(\Rightarrow-x=6-4x\)
\(\Rightarrow-x+4x=6\)
\(\Rightarrow3x=6\Rightarrow x=2\) (t/m)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{6}{5}\\x=2\end{matrix}\right.\).
b) \(\left|x-1\right|=3x+2\)
+) TH1: \(x-1\ge0\Rightarrow x\ge1\)
\(x-1=3x+2\)
\(\Rightarrow x-3x=1+2\)
\(\Rightarrow-2x=3\)
\(\Rightarrow x=\dfrac{-3}{2}\) (loại)
+) TH2: \(x-1< 0\Rightarrow x< 1\)
\(-x+1=3x+ 2\)
\(\Rightarrow-x-3x=-1+2\)
\(\Rightarrow-4x=1\)
\(\Rightarrow x=\dfrac{-1}{4}\) (t/m)
Vậy \(x=-\dfrac{1}{4}.\)
c) Tương tự.
a) Vì \(\left|\dfrac{1}{2}x\right|\ge0\)
\(\Leftrightarrow3-2x\ge0\)
Mà \(\left|\dfrac{1}{2}x\right|=3-2x\)
\(\Leftrightarrow\dfrac{1}{2}x=3-2x\)
\(\Leftrightarrow0,5x=3-2x\)
\(\Leftrightarrow0,5x+2x=3\)
\(\Leftrightarrow2,5x=3\)
\(\Leftrightarrow x=\dfrac{3}{2,5}\)
Vậy ................
b) Vì \(\left|x-1\right|\ge0\)
Mà \(\left|x-1\right|=3x+2\)
\(\Leftrightarrow3x+2\ge0\)
\(\Leftrightarrow x-1=3x+2\)
\(\Leftrightarrow x-3x=2+1\)
\(\Leftrightarrow-2x=3\)
\(\Leftrightarrow x=-\dfrac{3}{2}\left(tm\right)\)
Vậy ....................
c) Vì \(\left|5x\right|\ge0\)
Mà \(\left|5x\right|=x+12\)
\(\Leftrightarrow x+12\ge0\)
\(\Leftrightarrow5x=x+12\)
\(\Leftrightarrow5x-x=12\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy ............
Bà tag tui ko có trúng :v
a) \(\left|\dfrac{1}{2}x\right|=3-2x\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=3-2x\left(x\ge0\right)\\\dfrac{1}{2}x=2x-3\left(x< 0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+2x=3\\\dfrac{1}{2}x-2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{5}{2}x=3\\-\dfrac{3}{2}x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,2\\x=2\end{matrix}\right.\)
\(\Rightarrow x=1,2\)
các bài khác tương tự nhé !!
