Respuesta :

Step-by-step explanation:

Let the larger integer be x

Hence, the smaller integer = x-2

Hence,

Reciprocal of the x= 1/x

Reciprocal of x-2= 1/ x-2

Hence. their sum=

[tex] \frac{1}{x} + \frac{1}{x - 2} \\ = \frac{x - 2 + x}{x(x - 2)} \\ = \frac{2x - 2}{ {x}^{2} - 2x} [/tex]

The sum of the reciprocals of two consecutive even integers if the larger integer is [tex]x[/tex] is [tex]\frac{2}{x}[/tex].

For two consecutive even integers, if the larger integer is [tex]x[/tex], then the smaller one is [tex]x-2[/tex].

Their reciprocals will be [tex]\frac{1}{x}[/tex] and [tex]\frac{1}{x-2}[/tex].

Now, the sum of the reciprocals is [tex]\frac{1}{x} + \frac{1}{x-2}[/tex]

[tex]= \frac{1}{x} + \frac{1}{x-2}[/tex]

[tex]= \frac{x-2+x}{x(x-2)}[/tex]

[tex]= \frac{2x-2}{x(x-2)}[/tex]

[tex]= \frac{2(x-2)}{x(x-2)}[/tex]

[tex]= \frac{2}{x}[/tex]

So, the sum of the reciprocals of two consecutive even integers if the larger integer is [tex]x[/tex] is [tex]\frac{2}{x}[/tex].

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