Đặt \(A=2\left|2x-1\right|-3\left|2x+3\right|\)
TH1 : \(x< -\frac{3}{2};\) ta có :
\(A=2.\left(1-2x\right)-3.\left[-\left(2x+3\right)\right]\)
\(=2-4x+6x+9\)
\(=2x+11\)
TH2 : \(-\frac{3}{2}\le x< \frac{1}{2};\)ta có :
\(A=2\left(1-2x\right)-3\left(2x+3\right)\)
\(=2-4x-6x-9\)
\(=-10x-7\)
TH3 : \(x\ge\frac{1}{2};\)ta có :
\(A=2\left(2x-1\right)-3\left(2x+3\right)\)
\(=4x-2-6x-9\)
\(=-2x-11\)
+ Với \(x< -\frac{3}{2}\) thì |2x - 1| = 1 - 2x; |2x + 3| = -(2x + 3) = -2x - 3
Ta có: 2.|2x - 1| - 3.|2x - 3| = 2.(1 - 2x) - 3.(-2x - 3)
= 2 - 4x + 6x + 9
= 2x + 11
+ Với \(-\frac{3}{2}\le x< \frac{1}{2}\) thì |2x - 1| = 1 - 2x; |2x + 3| = 2x + 3
Ta có: 2.|2x - 1| - 3.|2x + 3| = 2.(1 - 2x) - 3.(2x + 3)
= 2 - 4x - 6x - 9
= -10x - 7
+ Với \(x>\frac{1}{2}\) thì |2x - 1| = 2x - 1; |2x + 3| = 2x + 3
Ta có: 2.|2x - 1| - 3.|2x + 3| = 2.(2x - 1) - 3.(2x + 3)
= 4x - 2 - 6x - 9
= -2x - 11
Đúng thì thôi c` sai thì cấm chửi
