Respuesta :
1. The denominator for the first one factors out to (x+2)(x+2). You get this by taking the coefficient of "4" and finding mult's that if adding together would give you the middle x term which is +4, then you would have x^2+2x+2x+4, you would use factor by grouping,(x^2+2x)(2x+4). Pull out the x from the first group, than a 2 from the second group. This will leave you with (x+2)(x+2). Now, the second part is "perfect squares," it is simply (2-x)(2+x). Now, with both of those done you come up with the final answer: (x+2)(2-x)is you LCD. If you are asking why not the (2+x), it is because that one is exact same as the (x+2), just written different but still the same value.
2. x/x^3+y^3, x/x+y: The first one is "perfect cubes" so it is the same as (x+y)^3, which simply means (x+y)(x+y)(x+y). Your second fraction is (x+y), they are the same so your LCD is (x+y).
3. 6/(a^2-7a+6): First you must factor the denominator, just like problem #1 but the coeffienct is now a "+6" and you have to find mult's of 6 that add up to a "-7." I know that two negatives = positive. -1*-6 give me my positive multiple and my -7 when added together so the after factoring by grouping you get: (a-1)(a-6). Now that you have factored your denominator, you multiply both the top and bottom by the LCD that was given (x-2)(x-1)(x-6). The denominator (x-1)(x-6) is cancelled out by the same two LCD's and the only group that is left is (x-2). This is the only group that is left for you to mult the top with. So, now you only have 6(a + 6) left, mult those through and get 6a+36, which is your final answer.
4. 2/(x-2): This one is simple, you just take the LCD given and mult it with (x-2). This leaves you with (x - 1)(x - 6) left and you mult these with the numerader for: 2(x - 1)(x - 6). Make sure you mult through to get your final answer. You need to finish this one by mult'g 2(x-1)(x-6), then you will have your final answer.
5. x-1/(x^2-x-20): Factor out the denominator like you did on the first examples then you should come up with (x-5)(x+4) for your denominator. Then mult it with the LCD (x - 5)(x + 4)(x + 3). The (x - 5)(x + 4) reduce to 1, there like before they are no longer there so all you have left is (x+3) to mult with the numerator and come up with your final answer.
2. x/x^3+y^3, x/x+y: The first one is "perfect cubes" so it is the same as (x+y)^3, which simply means (x+y)(x+y)(x+y). Your second fraction is (x+y), they are the same so your LCD is (x+y).
3. 6/(a^2-7a+6): First you must factor the denominator, just like problem #1 but the coeffienct is now a "+6" and you have to find mult's of 6 that add up to a "-7." I know that two negatives = positive. -1*-6 give me my positive multiple and my -7 when added together so the after factoring by grouping you get: (a-1)(a-6). Now that you have factored your denominator, you multiply both the top and bottom by the LCD that was given (x-2)(x-1)(x-6). The denominator (x-1)(x-6) is cancelled out by the same two LCD's and the only group that is left is (x-2). This is the only group that is left for you to mult the top with. So, now you only have 6(a + 6) left, mult those through and get 6a+36, which is your final answer.
4. 2/(x-2): This one is simple, you just take the LCD given and mult it with (x-2). This leaves you with (x - 1)(x - 6) left and you mult these with the numerader for: 2(x - 1)(x - 6). Make sure you mult through to get your final answer. You need to finish this one by mult'g 2(x-1)(x-6), then you will have your final answer.
5. x-1/(x^2-x-20): Factor out the denominator like you did on the first examples then you should come up with (x-5)(x+4) for your denominator. Then mult it with the LCD (x - 5)(x + 4)(x + 3). The (x - 5)(x + 4) reduce to 1, there like before they are no longer there so all you have left is (x+3) to mult with the numerator and come up with your final answer.