+ Nếu a < \(\frac{3}{2}\) thì |2x - 3| = 3 - 2x; |2 - x| = 2 - x
Ta có: (3 - 2x) - x = 2 - x
-> 3 - 3x = 2 - x
=> 3 - 2 = -x + 3x
=> 1 = 2x
\(\Rightarrow x=\frac{1}{2}\), thỏa mãn \(x< \frac{3}{2}\)
+ Nếu \(\frac{3}{2}\le x\le2\) thì |2x - 3| = 2x - 3; |2 - x| = 2 - x
Ta có: (2x - 3) - x = 2 - x
=> x - 3 = 2 - x
=> x + x = 2 + 3
=> 2x = 5
\(\Rightarrow x=\frac{5}{2}\), không thỏa mãn \(\frac{3}{2}\le x< 2\)
+ Nếu \(2< x\) thì |2x - 3| = 2x - 3; |2 - x| = x - 2
Ta có: (2x - 3) - x = x - 2
=> x - 3 = x - 2, vô lý
Vậy \(x=\frac{1}{2}\)
do |2x+3|=x+2 nên
=>*2x+3=x+2 hoặc *2x+3=-(x+2)
=>2x-x=2-3 =>2x+x=-2-3
=>x=-1 =>3x=-5
=>x=\(-\dfrac{5}{3}\)
/2x+3/-2/4-x/=5
<=> |2x+3|-2|4-x|=|2x+3|-2|x-4|
=> |2x+3|-2|x-4|=5
=>x= 5/2
