Hi I need help with this exponent problem.
(x^-1 + y^-1)^-2
thank you
I assume it is to simplify.
1/(1/x + 1/y)2
(xy)2/(y+x)2
remember this: (a+b)^2=(a+b)(a+b) so in
the denominator this happened
(xy)(1/x+1/y)(xy)(1/x+1/y)
(y+x)^2
thank you very much
_____1_______
(1/x + 1/y)^2
_________1_________
(1/x^2 + 2/xy + 1/y^2)
______x^2_y^2______
(y^2 + 2xy +x^2)
or
_x^2_y^2_
(x+y)^2
Of course! To solve the exponent problem (x^-1 + y^-1)^-2, let's break it down step by step.
Step 1: Simplify the expression inside the parentheses.
The expression inside the parentheses consists of two terms, both with negative exponents:
(x^-1 + y^-1)
To simplify this, we can rewrite it as a single fraction:
(x^-1 + y^-1) = 1/x + 1/y
Step 2: Square the expression inside the parentheses.
To square the expression 1/x + 1/y, we simply multiply it by itself:
(1/x + 1/y)^2 = (1/x + 1/y) * (1/x + 1/y)
Step 3: Expand and simplify.
To expand the squared expression, we use the distributive property:
(1/x + 1/y) * (1/x + 1/y) = (1/x)*(1/x) + (1/x)*(1/y) + (1/y)*(1/x) + (1/y)*(1/y)
Multiplying the numerators and denominators gives:
= 1/(x^2) + 1/(xy) + 1/(yx) + 1/(y^2)
Since (xy) is the same as (yx), we can simplify this to:
= 1/(x^2) + 2/(xy) + 1/(y^2)
So, the simplified expression of (x^-1 + y^-1)^-2 is:
1/(x^2) + 2/(xy) + 1/(y^2)
I hope this helps! Let me know if you have any further questions.