a2-2ab+b2/ a-b?

Answer 1

(a - b) (a - b) / (a - b) = (a - b)

Answer 2

I suppose you mean (a^2 - 2ab + b^2) / (a - b).

(a^2 - 2ab + b^2) / (a - b).
= (a - b)^2 / (a - b)
= a - b
[Only if a - b ≠ 0 since we divided numerator and denominator by (a - b).]

Answer 3

(a^2-2ab+b^2)/(a-b)=?
well, a^2+2ab+b^2 is the answer to (a+b)^2, and in this case, we have -2ab, so its (a-b)^2.
thus, the problem becomes:
(a-b)^2/(a-b)
or:
(a-b)(a-b)/(a-b)
one of the (a-b)s on the top cancels the one on the bottom. thus, we are left with:
(a-b)

Answer 4

a^ - 2ab + b^2 / a - b
factor the numerator:

(a - b)(a - b) / (a - b)

The (a - b) in the numerator and the (a - b) in the denominator cancel each other out, leaving the answer of (a -b)

Answer 5

lol, the final clause replaced the complete question. that's real while A and B are scalars. that's not real while A and B are vectors, matrices, and so on. In a greater compact assertion, right here may be suggested approximately tensors: Rank 0 real Else fake keep in mind that a tensor of rank 0 is a scalar, rank a million is a vector, rank 2 is regardless of, and so on. The question is somewhat too uninteresting for me to p.c. to kind out a information. See your textbook, it is going to probably additionally be in a sophisticated calculus I textbook interior the introductory chapters.

Answer 6

You need to factor the numerator first.

a^2 - 2ab + b^2
(a - b)(a - b)

When you put that back over your denominator, one factor will cancel.

[(a - b)(a - b)]/(a - b)

You are left with (a - b)

Answer 7

(a^2 -- 2ab + b^2) / (a -- b)
= (a -- b)(a -- b) / (a -- b)
= (a -- b)

Answer 8

1 apple add 1 banana cant be 2 banapples so it has to be 1 apple and 1 banana... that is all algebra is...once you understand that,its all elementary.

Answer 9

a-b