当x+2≥0 且 2x-3≤0 即:-2≤x≤3/2时 |x+2|+|2x-3|=x+2+(3-2x)=5-x
∴-2≤x≤3/2
(x-4)√(4x²-12x+9)+(1+2x)√(x²+4x+4)
=(x-4)︱2x-3︱+(1+2x)︱x+2︱ ∵-2≤x≤3/2
=(x-4)(3-2x)+(1+2x)(x+2)
=16x-10
当x+2≥0 且 2x-3≤0 即:-2≤x≤3/2时 |x+2|+|2x-3|=x+2+(3-2x)=5-x
∴-2≤x≤3/2
(x-4)√(4x²-12x+9)+(1+2x)√(x²+4x+4)
=(x-4)︱2x-3︱+(1+2x)︱x+2︱ ∵-2≤x≤3/2
=(x-4)(3-2x)+(1+2x)(x+2)
=16x-10