a, +) Xét \(x\ge2,5\) có:
\(x-1,5+x-2,5=0\)
\(\Leftrightarrow2x=4\Leftrightarrow x=2\) ( không t/m )
+) Xét \(1,5\le x< 2,5\) có:
\(x-1,5+2,5-x=0\)
\(\Leftrightarrow1=0\) ( ko t/m )
+) Xét x < 1,5 có:
\(1,5-x+2,5-x=0\)
\(\Leftrightarrow2x=4\Leftrightarrow x=2\) ( ko t/m )
Vậy không có giá trị x thỏa mãn
b, \(\left\{{}\begin{matrix}\left|x+3\right|\ge0\\\left|x+1\right|\ge0\end{matrix}\right.\Leftrightarrow\left|x+3\right|+\left|x+1\right|\ge0\)
\(\Leftrightarrow3x\ge0\Leftrightarrow x\ge0\)
\(\Leftrightarrow x+3+x+1=3x\)
\(\Leftrightarrow x=4\)
Vậy x = 4
c, \(\left|x-7\right|=1-2x\)
+) Xét \(x\ge7\) có:
\(x-7=1-2x\Leftrightarrow3x=8\Leftrightarrow x=\dfrac{8}{3}\)( ko t/m )
+) Xét x < 7 có:
\(7-x=1-2x\Leftrightarrow x=-6\) ( t/m )
Vậy x = -6
\(\left|x-1,5\right|+\left|2,5-x\right|=0\)
\(\left|x-1,5\right|\ge0\) \(\left|2,5-x\right|\ge0\)
Dấu "=" xảy ra khi
\(\left|x-1,5\right|=0\Rightarrow x-1,5=0\Rightarrow x=1,5\)
\(\left|2,5-x\right|=0\Rightarrow2,5-x=0\Rightarrow x=2,5\)
\(\left|x+3\right|+\left|x+1\right|=3x\)
\(\left|x+3\right|\ge0;\left|x+1\right|\ge0\)
\(\Leftrightarrow\left|x+3\right|+\left|x+1\right|\ge0\)
\(\Rightarrow x+3+x+1=3x\)
\(2x+4=3x\)
\(x=4\)
