Question

Need ASAP please and thank you! :)

Answer 1

B.
`(x^2+3)/(\left(x-1\right)\left(x-3\right))`

Explanation:

Will provide explanation later since you are in a hurry

1. Find the LCM of the two denominators: x-1 and x -3

This is (x-1)(x-3)

2. Multiply each numerator by the same amount needed to multiply its​corresponding denominator to turn it into the LCM (x−1)(x−3)

`\mathrm{For}\:(x-3)/(x-1):\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:x-3`

`(x-3)/(x-1) = (\left(x-3\right)\left(x-3\right))/(\left(x-1\right)\left(x-3\right)) = (\left(x-3\right)^2)/(\left(x-1\right)\left(x-3\right))`

`\mathrm{For}\:(6)/(x-3):\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:x-1`

`(6)/(x-3) = (6\left(x-1\right))/(\left(x-3\right)\left(x-1\right)) = (6\left(x-1\right))/(\left(x-1\right)\left(x-3\right))`

`2. \mathrm{Simplify\:}\left(x-3\right)^2+6\left(x-1\right)`

`\left(x-3\right)^2 = x^2 - 6x + 9\\\\\\3. \; \text{Expand }6\left(x-1\right)\;6\left(x-1\right) = 6x-6\\\\`

4. Since the denominators are the same in both terms, we can add the numerators and use the common denominator as the denominator for the result

• Adding numerators derived from steps 2 and 3 above we get
  
  `x^2-6x+9+6x-6 = x^2+3`

• Dividing by the common denominator (x-1)(x-3) gives the result as
  
  `(x^2+3)/(\left(x-1\right)\left(x-3\right))`
  

Answer 2

While adding two Fractions first we find the LCM of Denominators,

The LCM of x-1 and x-3 is (x-1)(x-3)

now, we perform the calculation as ,

`{(x-3)/(x-1)} + {(6)/(x-3)}`

`{((x-3)²+6(x-1))/((x-1)(x-3))}`

`{(x²+9-6x +6x-6)/((x-1)(x-3))}`

`{(x²+3)/((x-1)(x-3))}`

Hence option B is the answer