中一Maths

100÷[(___-1+3x3-___)x(8+___-10-___+5)]=√10000

100÷[(___-1+3x3-___)x(8+___-10-___+5)]=+/- 100
將空格當做ABCD
[(_A__-1+3x3-_B__)x(8+__C_-10-__D_+5)]=+/-1

(A-1+3x3-B)x(8+C-10-D+5)=+/-1

(A+8-B)x(C-D+3)=+/-1
(A+8-B)x(C-D+3)=+1 OR (A+8-B)x(C-D+3)= -1
先解 +1
(A+8-B)=1 and (C-D+3)=1
A=B-7 C=D-2
so A 可以係2 B=9 C=6 D=4 ...
當右邊係 = -1
(A+8-B)x(C-D+3)= -1
(A+8-B)=1 (C-D+3)= -1 OR (A+8-B)= -1 (C-D+3)= 1
A=B-7 C=D-4 A=B-9 C=D-2
so A =2 B=9 C=6 D=2 A=1 B=10 C=6 D=4 ...
有好多種可能.... 希望幫到你喇 ^^

100/[(1-1+3*3-8)*(8+2-10-4+5)]