中四amaths 絕對值既數.. 請幫忙解答,THX

1
| X-2 | = | X^2 - 4 |
| X-2 | = | (X+2)(X-2) |
X+2=1 or X+2=-1
X=-1 or X=-3
since when X=2 each side is equal to 0
ANS: X=-3,-1,2
#2 ( X - 3 )^2 - | X - 3 | - 12 = 0
let y=X-3
then the equation become
y^2-|y|-12=0
when y>=0
y^2-y-12=0
(y-4)(y+3)=0
y=4 or y=-3 (reject)
so X-3=4 ,X=7
when y<0
y^2+y-12=0
(y+4)(y-3)=0
y=-4 or y=3 (reject)
so X-3=-4 ,X=-1
ANS:X=-1,7

#1

| X-2 | = | X^2 - 4 |

x-2=x^2-4 或 x-2=-(x^2-4)
0=x^2-x-4+2 或 0=-x^2-x+4+2
0=x^2-x-2 或 0=-x^2-x+6
0=(x-2)(x+1) 或 0=(-x+2)(x+3)
所以x-2=0,x+1=0,-x+2=0或x+3=0
x=2,-1或-3

第一題上面答左


2 .


(x-3)^2-|x-3|-12=0

|x-3|^2-|x-3|-12=0

(|x-3| +3 ) (|x-3|-4) = 0

For |x-3| = -3

since |x|=x, &lt; 0
so, -3 is rejected.

For |x-3|=4
x-3 =+4 or -4
x = 7 or x = -1