a) \(\left|2x-3\right|-x=\left|2-x\right|\)
\(\Leftrightarrow\left|2x-3\right|-x-\left|2-x\right|=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3-x-\left(2-x\right)=0\left(đk:2x-3\ge0;2-x\ge0\right)\\-\left(2x-3\right)-x\left(2-x\right)=0\left(đk:2x-3< 0;2-x\ge0\right)\\2x-3-x-\left(-\left(2-x\right)\right)=0\left(đk:2x-3\ge0;2-x< 0\right)\\-\left(2x-3\right)-x-\left(-\left(2-x\right)\right)=0\left(đk:2x-3< 0;2-x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(đk:x\ge\dfrac{3}{2};x\le2\right)\\x=\dfrac{1}{2}\left(đk:x< \dfrac{3}{2};x\le2\right)\\x\in\varnothing\left(đk:x\ge\dfrac{3}{2};x>2\right)\\x=\dfrac{5}{4}\left(đk:x< \dfrac{3}{2};x>2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=\dfrac{1}{2}\\x\in\varnothing\\x\in\varnothing\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
a, Đề theo mình nghĩa là -|2-x|
b, \(\left|x+3\right|+\left|x+1\right|=3x\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left|x+3\right|\ge0;\left|x+1\right|\ge0\)
\(\Rightarrow\left|x+3\right|+\left|x+1\right|\ge0\)
mà \(\left|x+3\right|+\left|x+1\right|=3x\)
nên \(3x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+3\right|=x+3;\left|x+1\right|=x+1\)
Từ đó:\(\left|x+3\right|+\left|x+1\right|=3x\)
\(\Rightarrow x+3+x+1=3x\)
\(\Rightarrow2x+4=3x\Rightarrow3x-2x=4\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
Chúc bạn học tốt!!!
a, \(\left|2x-3\right|-x=\left|2-x\right|\)
\(\Leftrightarrow\left|2x-3\right|-\left|x-2\right|=x\)
+) Xét \(x\ge2\) có:
\(2x-3-x+2=x\)
\(\Leftrightarrow x-1=x\Leftrightarrow0=1\) ( vô lí )
+) Xét \(1,5\le x< 2\) có:
\(2x-3-2+x=x\)
\(\Leftrightarrow2x=5\)
\(\Leftrightarrow x=\dfrac{5}{2}\) ( không t/m )
+) Xét \(x< 1,5\) có:
\(3-2x-2+x=x\)
\(\Leftrightarrow1-x=x\)
\(\Leftrightarrow2x=1\Leftrightarrow x=0,5\) ( t/m )
Vậy x = 0,5
b, Ta có: \(\left\{{}\begin{matrix}\left|x+3\right|\ge0\\\left|x+1\right|\ge0\end{matrix}\right.\Leftrightarrow\left|x+3\right|+\left|x+1\right|\ge0\)
\(\Leftrightarrow3x\ge0\Leftrightarrow x\ge0\)
\(\Leftrightarrow x+3+x+1=3x\)
\(\Leftrightarrow2x+4=3x\Leftrightarrow x=4\)
Vậy x = 4
\(\left|x+3\right|+\left|x+1\right|=3x\)
\(\left|x+3\right|\ge0\)
\(\left|x+1\right|\ge0\)
\(\Leftrightarrow\left|x+3\right|+\left|x+1\right|\ge0\)
\(\Leftrightarrow x+3+x+1=3x\)
\(2x+4=3x\)
\(\Leftrightarrow x=4\)
