+0  
 
0
104
3
avatar

Select the exclusion that fits best with this problem.

 

a^3-b^3/a^3b^3+a^2b^4+ab^5=

 

x,y,z=0

x=0

1+y+y^2, y=0, 1-y=0

2x-1=0, 4x^2+2x+1=0

a, b= 0, a^2+ab+b^2=0

 May 1, 2023
 #1
avatar+137 
0

x,y,z=0

Step-by-step explanation:

For this case we have the following expression:
 3x ^ 2y ^ 2-6xz / 6x ^ 2yz
 From here, we must exclude all values that make the denominator equal to zero.
 We have then:
 6x ^ 2yz = 0
 Therefore, the results are:
 x = 0
 y = 0
 z = 0
 Answer:
 The exclusion that fits best with this problem is:
 x, y, z = 0

 May 1, 2023
 #2
avatar
0

I don't get how you got that answer. The variables are A and B. 

Guest May 1, 2023
 #3
avatar+137 
0

Well ' I ' concluded because from looking at the problem:

a^3-b^3 / a^3b^3+a^2b^4+ab^5 it is a fraction correct?

And from the fraction the calue of variable of a expression are excluded for which expression does not exist, where denominator is going to be zero those value going to be excluded. So making the denominator ( NOT ' 0 ' ) so like in The question best fits the problem: 

I need to find point where it become zero we put 6x²yz equal to 0

6x²yz = 0

 

Making the new problem of:x² = 0  or y = 0 or z = 0

x² = 0  or y = 0 or z = 0

 

and concealing the problem making option a correct.

acyclics  May 1, 2023

1 Online Users