radnorgardens
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 Dec 2, 2014
 Messages
 26
Question:
. . . . .\(\displaystyle \mbox{(e) }\, \dfrac{5}{x\, (x\, +\, 1)}\, \, \dfrac{2}{x}\, +\, \dfrac{3}{x\, +\, 1}\)
I'm getting myself confused with this one.
1) I multiplied all the denominators, so they all became x(x+1)x(x+1).
2) Then 5x(x+1)  2x(x+1)(x+1) + 3x(x+1)x
3) Becomes, 5x[SUP]2[/SUP] + 5x  2x[SUP]3[/SUP]  4x[SUP]2[/SUP]  2x + 3x[SUP]3[/SUP] + 3x[SUP]2[/SUP]
4) Simplified, 4x[SUP]2 [/SUP]+ 3x + x[SUP]3 [/SUP]/ x(x+1)x(x+1).
The answer is x + 3 / x(x+1). I don't know how to get this.
Appreciate your help all.
Thxs.
. . . . .\(\displaystyle \mbox{(e) }\, \dfrac{5}{x\, (x\, +\, 1)}\, \, \dfrac{2}{x}\, +\, \dfrac{3}{x\, +\, 1}\)
I'm getting myself confused with this one.
1) I multiplied all the denominators, so they all became x(x+1)x(x+1).
2) Then 5x(x+1)  2x(x+1)(x+1) + 3x(x+1)x
3) Becomes, 5x[SUP]2[/SUP] + 5x  2x[SUP]3[/SUP]  4x[SUP]2[/SUP]  2x + 3x[SUP]3[/SUP] + 3x[SUP]2[/SUP]
4) Simplified, 4x[SUP]2 [/SUP]+ 3x + x[SUP]3 [/SUP]/ x(x+1)x(x+1).
The answer is x + 3 / x(x+1). I don't know how to get this.
Appreciate your help all.
Thxs.
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