Given that abc is less than 0 and a+b+c is greater than 0, when x=a is the absolute value divided by a+b and b+c is the absolute value divided by c, the value of x to the 19th power +92x+2 is obtained

Given that abc is less than 0 and a+b+c is greater than 0, when x=a is the absolute value divided by a+b and b+c is the absolute value divided by c, the value of x to the 19th power +92x+2 is obtained

X=|a a b b c c|
Abc <0, two positive one negative or three negative in three numbers
A + b + c >0, so three numbers must be two positive and one negative.
So, x=1+1+(-1)=1
19Th power of x +92 x +2=1+92+2=95

X=|a a b b c c|
Abc <0, two positive one negative or three negative in three numbers
A + b + c >0, so three numbers must be positive and negative.
So, x=1+1+(-1)=1
19Th power of x +92 x +2=1+92+2=95

Negative number minus negative number such as -1-(-1)=? (-1)1St power +(-1)2nd power +(-1)3rd power +...+(-1)2006th power =?

(-1)1St power +(-1)2nd power +(-1)3rd power +...+(-1)2006 power
=0
The odd power of (-1) is -1 and the even power is 1
The original formula starts with odd power and ends with even power, so the number of -1 and 1 is just equal, so the result is 0