Question

代数式的简化

Answer 1

更新1:

Q2. 4x-3 ------ -(减) 8 3x+1 ------ 4 =??????

1. 最小公倍数 L.C.M (lowest mon multiple) L.C.M. of 6 and 4 系 12 (3x+1)/6 - (2x+3)/4 =(6x+2)/12 - (6x+9)/12 =(6x + 2 - 6x - 9)/12 =7/12 2. L.C.M. of 8 and 4 系8 (4x-3)/8 - (3x+1)/4 =(4x-3)/8 - (6x+2)/8 =(4x - 3 - 6x - 2)/8 = (-2x - 5)/8
3x+1)/6－(2x+3)/4=???? Sol [6
4]=12 A=(3x+1)/6-(2x+3)/4 12A=12*[(3x+1)/6-(2x+3)/4] =2*(3x+1)－3*(2x+3) =6x+2-6x－9 =－7 A=－7/12 (3x+1)/6-(2x+3)/4=－7/12 Q.2 不知