三題數學,﹝己有問案或過程﹞

1.若：a^2=-﹝4﹞^2，b^2=4^3,則│a+b│的值是_____.﹝12或4﹞
請解釋並寫過程
a^2=-﹝4﹞^2 that means a^2=16, a=+4 or -4
b^2=4^3 that means b^2=64, b=+8 or -8
if a=+4, b=+8, │a+b│= │4+8│=12;
if a=+4, b=-8, │a+b│= │4-8│=4;
if a=-4, b=+8, │a+b│= │-4+8│=4;
if a=-4, b=-8, │a+b│= │-4-8│=12;

2.若：│a│=4， √b^2=3,且│a+b│=-a-b,則a-b的值是___.﹝-1/-7﹞
請解釋並寫過程
│a│=4 that means a=+4 or -4
√b^2=3 that means b=+3 or 3
coz│a+b│=-a-b =-(a+b), so a+b<0, a<-b;
so a≠+4, just a=-4;
if a=-4, b=+3, a-b=-4-3=-7;
if a=-4, b=-3, a-b=-4+3=-1.

3.﹝x^2+3x+3﹞/﹝x^2+3x+2﹞-﹝x^2+3x-2﹞/﹝4-x^2﹞+﹝9-2x-2x^2﹞/﹝x^2-x-2﹞
解:原式=1+1/﹝x^2+3x+2﹞+1+﹝3x+2﹞/﹝x^2-4﹞-2-﹝4x-5﹞/﹝x^2-x-2﹞
↑尼一步點得的?
I will tell you later.

2007-08-06 00:55:16 補充：
3.解:原式=[(x^2+3x+2)+1] / (x^2+3x+2) + [(x^2-4)+(3x+2)] / (x^2-4)+ [-2(x^2-x-2)-(4x-5)] / (x^2-x-2) = (x^2+3x+2) / (x^2+3x+2) + 1/ (x^2+3x+2) + (x^2-4) / (x^2-4) + (3x+2) / (x^2-4) - 2(x^2-x-2) / (x^2-x-2) - (4x-5) / (x^2-x-2) =1+1/﹝x^2+3x+2﹞+1+﹝3x+2﹞/﹝x^2-4﹞-2-﹝4x-5﹞/﹝x^2-x-2﹞

1.
a2次方=-4 gei2次方.
姐係 a2次方=16
姐係a=4

b2次方=4gei3次方
姐係b2次方=64
咁b就=8
咁a+b=12