The original formula = (1-1/2) × (1+1/2 )× (1-1/3 )× (1+/kloc). ×( 1+ 1/99)×( 1- 1/ 100)×( 1+ 1/ 100)

= 1/2×3/2×2/3×4/3×3/4×5/4×...×98/99× 100/99×99/ 100× 10 1/ 100

= 1/2× 10 1/ 100= 10 1/200

2. Question ① is that a square is missing, which should be: b 2+2ab = c 2+2ac.

b^2+2ab-c^2-2ac=0

(b+c)(b-c)+2a(b-c)=0

(b-c)(b+c+2a)=0

∴ b = c, isosceles triangle

②a^2-b^2+c^2-2ac

=(a^2+c^2-2ac)-b^2

=(a-c)^2-b^2

=(a-c+b)(a-c-b)

The sum of two sides of a triangle is greater than the third side, and the difference between the two sides is smaller than the third side.

a-c+b>0，a-c-b