✔ 最佳答案
1. 0<a<b<1 => 0<a^2<b^2<1(A)錯
0<a<b<1 =>1/a>1/b>1 (B)對
0<a<b<1 =>0<a+b<1+1=2 (C)對
0<a<b<1 =>0<ab<1 =>ab-1<0 (D)對
2. (x-3)+(26-x)>2x-7
23>2x-7
15>x----------------------
(x-3)+(2x-7)>(26-x)
3x-10>26-x
4x>36
x>9-----------------------
(2x-7)+(26-x)>(x-3)
x+19>x-3
22>0
So
15>x>9
14>=x>=10
14-10+1=5
(B)