Quick math problem - Training Enrollment Network

Quick math problem

Solution: There are four situations:

(1) When a, b and c are all positive numbers,

a/|a|+b/|b|+c/|c|+abc/|abc|

= 1+ 1+ 1+ 1

② When the numbers A, B and C are all negative,

a/|a|+b/|b|+c/|c|+abc/|abc|

=- 1- 1- 1- 1

=-4

(3) When there are two positive numbers and one negative number in the three numbers A, B and C,

a/|a|+b/|b|+c/|c|+abc/|abc|

= 1+ 1- 1- 1

(4) When there are two negative numbers and one positive number in the three numbers A, B and C,

a/|a|+b/|b|+c/|c|+abc/|abc|

=- 1- 1+ 1+ 1

So a/|a|+b/|b|+c/|c|+abc/|abc| has three values, the maximum value is 4, the minimum value is -4, and the other value is 0.

So m=4 and n=-4.

Male (female) name

=4÷(-4-4)

=4÷(-8)

= - 1/2