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Question

Simplify $$\displaystyle \frac {1}{(x+1)(x+2)}+\frac {1}{(x+2)(x+3)}+\frac {1}{(x+2)(x+3)}$$


A
1(x+1)(x+3)
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B
2(x+1)(x+3)
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C
3(x+2)(x+3)
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D
3(x+1)(x+3)
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Solution

The correct option is D $$\dfrac {3}{(x+1)(x+3)}$$
$$\displaystyle\frac {1}{(x+1)(x+2)}+\frac {1}{(x+2)(x+3)}+\frac {1}{(x+3)(x+1)}=\frac {x+3+x+1+x+2}{(x+1)(x+2)(x+3)}$$

$$=\dfrac {3x+6}{(x+1)(x+2)(x+3)}$$

$$=\dfrac {3(x+2)}{(x+1)(x+2)(x+3)}=\dfrac {3}{(x+1)(x+3)}$$

Mathematics

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