Use complete sentences to describe the steps taken to simplify this problem. Make sure you include information about the new common denominator, the final simplified expression, and any restrictions (3/x)-(3/x+1)-(3/x+2)
\[2-x ^{2}/(x)(x+1)(x+2)\]
we have the exprssion 3/x-3/(x+1)-3/(x+2) take lcm of x, x+1 and x+2 which is x(x+1)(x+2) (3(x+1)(x+2)-3(x)(x+2)-3(x)(x+1))/(x(x+1)(x+2)) taking 3 common from numerator and expanding the brackets 3[x^2+3x+2-x^2-2x-x^2-x]/(x(x+1)(x+2) bringing terms of common power of x together 3[x^2-x^2-x^2+3x-2x-x+2]/(x(x+1)(x+2)) 3[-x^2+2]/(x(x+1)(x+2)) \[3(2-x^2)/(x(x+1)(x+2))\]
can you understanding this answer ?
so for this exersice you need using the same procese like in case of 1/2 +1/3 + 3/4 than you need getting the lcd for 2,3 and 4 - so how you can seeing that will be 12 so now you see that in case of 1/2 the denominator 2 till 12 you need multiply by 6 so multiply the 1 by 6 and in case of 1/3 you see that 3 till 12 you need multiply by 4 and in case of 3/4 you see that 4 till 12 you need multiply by 3 - so now after these multiplications of numerators check what will get like numerator finale so will be 6+4+3 13 ------- = ---- 12 12 - now i think that you will can doit the same in case of functions like x+1 or ... hope so much that is understandably sure ! good luck bye
so you need to calcule the denominator common for x and x+1 and x+2 so what will be x*(x+1)*(x+2) and hence you ned multiply the first numerator by (x+1)(x+2) and second numerator by x(x+2) and 3-rd numerator by x(x+1) so after this you make those calcules and will get (3(x+1)(x+2)-3x(x+2)-3x(x+1))/(x(x+1)(x+2)
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