(1/a+1/b)÷(a-²–b-²)
Answers
Step-by-step explanation:
So a^2 - b^2 = 1 which, when factored produces
(a-b)(a+b) = 1
Squaring, (a-b)^2 * (a+b)^2 = 1 and dividing by each term on the LHS we get two equations:
(a+b)^2 = 1/(a-b)^2
(a-b)^2 = 1/(a+b)^2
Now adding both equations,
1/(a-b)^2 + 1/(a+b)^2
= (a+b)^2 + (a-b)^2
= (a^2 + 2ab + b^2) + (a^2 - 2ab + b^2)
= 2(a^2 + b^2)
but from the first equation given, a^2 = b^2 + 1 so that
2(a^2 + b^2)
= 2(2a^2 + 1)
Now for any a in R, the smallest value 2a^2 + 1 can take on is when a = 0, and evaluating this expression at a = 0 produces 1 (2(0)^2 + 1 = 1).
Prove this by differentiating 2a^2 + 1 w.r.t a and equating it to 0, and then taking the second derivative w.r.t a and noting that it is positive, a = 0 is a minimum.
Therefore, since a^2 >= 0, 2a^2 + 1 >= 2(0)^2 + 1 = 1 implying that
2a^2 + 1 >= 2 for all a in R.
HOPE ITS HELP YOU
Knowing this, 2(2a^2 + 1) >= 2(1) = 2,
but 2(2a^2 + 1) = 1/(a-b)^2 + 1/(a+b)^2 which is the expression
Answer:
answer of this question is in picture
Step-by-step explanation:
hipe ! it's helps
