Question #bf791
1 Answer
The answer is (B)
Explanation:
The idea here is that you need to take your starting eexpressions
#1/(1/(x+2) + 1/(x+3))#
and do some algebraic manipulations to get it to math one of the four expressions given.
Focus on the denominator of the original fraction
#1/(x+2) + 1/(x+3)#
In order to be able to add these two fractions, you need them to have the same denominator. This means that you're going to have to multiply the first one by
#1/(x+2) * (x+3)/(x+3) + 1/(x+3) * (x+2)/(x+2)#
#(x+3)/((x+2)(x+3)) + (x+2)/((x+2)(x+3))#
Now you can add these fractions by adding their numerators
#( x+3 + x+2)/((x+2)(x+3)) = (2x + 5)/((x+2)(x+3))#
The original fraction can thus be written as
#1/((2x+5)/((x+2)(x+3))#
Now, you know that dividing a number by a fraction is equivalent to multiplying the number by the inverse of the fraction
#color(blue)(a/(b/c) = a * c/b)#
In your case,
#1/((2x+5)/((x+2)(x+3))) = 1 * ((x+2)(x+3))/(2x+5) = ((x+2)(x+3))/(2x+5)#
Finally, you can expand the parantheses of the numerator to get
#(x+2)(x+3) = x^2 + 2x + 3x + 6 = x^2 + 5x + 6#
This means that the expression will be equal to
#((x+2)(x+3))/(2x+5) = (x^2 + 5x + 6)/(2x+5)#
Therefore, you have
#1/(1/(x+2) + 1/(x+3)) = color(green)((x^2 + 5x + 6)/(2x+5))#