Question

Simplify completely..........​

Answer 1

Answer:

$\frac{x}{x-1}$

Step-by-step explanation:

$\frac{3x^2 - 1}{x^2 - 1} - \frac{2x + 1}{x + 1}$                                          $[ \ x^ 2- 1 = (x-1)(x + 1) \ ]$

$= \frac{3x^2 - 1}{(x-1)(x + 1 )} - \frac{2x + 1}{x + 1}\\\\=\frac{(3x^2 - 1)-(2x + 1)(x-1)}{(x + 1)(x-1)}$                        $[\ Taking \ LCM \ ]$

$= \frac{(3x^2 - 1)- (2x^2 - 2x + x - 1)}{(x+1)(x-1)}\\\\=\frac{3x^2 - 1 - 2x^2 + x + 1 }{(x+1)(x-1)}\\\\=\frac{x^2 +x}{(x+1)(x-1)}\\\\=\frac{x(x+1)}{(x+1)(x-1)}\\\\=\frac{x}{x-1}$

Finding LCM :

Example :

                 $\frac{1}{6} + \frac{1}{3}$

6 = 2  x   3

3 = 1   x   3

                $\frac{1}{2 \times 3} + \frac{1}{ 3}$ $= \frac{1}{2 \times 3} + \frac{1 \times 2}{ 3 \times 2}$  

[ To make the denominators same : the second fraction is multiplied and divided by 2 ]

Similarly :

 $(x^2 - 1 ) = (x -1)(x+1)\\\\(x + 1) = 1 \times (x + 1)$

Same rule we applied : multiplied the numerator and denominator of the second term with ( x - 1 )

Therefore the second term becomes ,

                                       $\frac{2x + 1}{x + 1} = \frac{(2x + 1)(x - 1)}{(x + 1)( x - 1)}$