# Expressed the following in the simplest form, \displaystyle \frac{(n-\frac{1 }{n }) }{(1+\frac{1...

## Question:

Expressed the following in the simplest form,

{eq}\displaystyle \frac{(n-\frac{1 }{n }) }{(1+\frac{1 }{n }) } {/eq} is equivalent to

{eq}1.)n-1 {/eq}

{eq}2.)n+1 {/eq}

{eq}3.)\frac{n-1 }{n+1} {/eq}

{eq}4.)n {/eq}

## Evaluating Simple Algebraic Expressions:

Algebraic expressions can be evaluated similarly as we evaluate simple mathematical expressions. The arithmetic operations of subtraction addition, multiplication, and division can be performed in same manner as we do them for simple numbers or fractions.

For the expression given in the question, the first step will be to simplify the numerator as well as the denominator.

Let's write the expression first...

{eq}\dfrac{n-\dfrac{1}{n}}{\dfrac{1}{n}+n} {/eq}.

On simplifying the numerator and the denominator, simultaneously we get to the following expression.

{eq}\dfrac{\dfrac{n^{2}-1}{n}}{\dfrac{n+1}{n}}=\dfrac{n^{2}-1}{n+1}=\dfrac{(n-1)(n+1)}{(n+1)}=\dfrac{n-1}{1}=n-1 {/eq}

So, {eq}(n-1) {/eq} will be the answer. 