( 1) ab-2b≠0
(2)△=4(b-a)^2-4(ab-2b)(2a-ab)=0
B 2+2ab+b 2+a 2b 2-2a 2b-2ab 2 = 0。
At the same time, a^2b^2 is imposed on both sides, which is simplification.
( 1/a+ 1/b)^2-2( 1/a+ 1/b)+ 1=0
( 1/a+ 1/b- 1)^2=0
( 1/a+ 1/b- 1)=0
1/a+ 1/b= 1
2) In 1 equation, △=2a+b-2* radical sign cd.
Delta = 2b+c-2 * radical sign ad in two equations.
Among the three equations, △=2c+d-2* radical sign ab
In the four equations, △=2d+b-2* radical bc.
Add up the delta of 1 3 and disassemble it, that is.
(AB+B under a-2* radical sign)+(CD+D under c-2 * radical sign)+A+C =
(radical a- radical B) 2+ (radical c+ radical d) 2+a+C.
So: (radical a- radical B) 2+ (radical c+ radical D) 2+A+C > 0.
At least two equations have unequal real roots.