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