[amaths]some ask about absolute values

for 0≦x≦1
-(x-1) = x-1
x=1

for x<0
-(x-1) = -x-1
-x+1 = -x-1
1=-1
which is not true
so x<0 has no solution

for x>1
x-1=x-1
1=1
which is always true
so the overall solution is x>= 1


通常二邊有個absolute sign + constant 都要分case(eg. | x-1 | = | x | -1 )
二邊有absolute but no constant or 得一邊有個absolute 都唔使分case, 分正負就得ga la

好似 solve
| x-1 | = | x | -1
就要分 cases

有三個情況
1) x <= 0
| x-1 | = -(x - 1)
| x | = -x
so, -(x - 1) = -x - 1

2) 0 < x <= 1
| x-1 | = -(x - 1)
| x | = x
so, -(x - 1) = x - 1

3) x > 1
| x-1 | = (x - 1)
| x | = x
so, (x - 1) = x - 1

就係要諗 x 係邊個範圍個 absolute value 可以拆得
上面例子, 有兩個 absolute values, 所以有三個情況

比多個簡單 d 既例子你睇,等你清楚 d
| x-3 | = 2
應該要分兩個情況啦
1) x <= 3,
| x-3 | = -( x-3 )

2) x > 3,
| x-3 | = x-3

如果你選其他情況, 如
x <= 5, | x-3 | 拆唔走, 因為佢有可能正,有可能負