Question

Check down the image that’s give below

Answer 1

Solution:

$\frac{3x}{ {x}^{2} + 6x + 9} + \frac{x + 3}{ {x}^{2} - 9 }$

In the first fraction, we have to factorise the denominator using (a + b)² = a² + 2ab + b². And in the second fraction, we have to factorise the denominator using a² - b² = (a - b)(a + b) identity.

$= \frac{3x}{ {(x)}^{2} + 2(x)(3) + ( {3)}^{2} } + \frac{x + 3}{(x)^{2} - (3)^{2} } \\ = \frac{3x}{(x + 3) ^{2} } + \frac{x + 3}{(x + 3)(x - 3)}$

From the second fraction, cancel out from both sides (x + 3), the we get:

$= \frac{3x}{(x + 3) ^{2} } + \frac{1}{(x - 3)} \\ = \frac{3x(x - 3) + 1(x + 3) ^{2} }{(x + 3)^{2} (x - 3)} \\ = \frac{3x ^{2} - 9 x+ {x}^{2} + 6x + 9}{ {(x + 3)}^{2} (x - 3)} \\ = \frac{ {4x}^{2} - 3x + 9}{(x + 3)(x + 3)(x - 3)}$

Answer:

$\frac{ {4x}^{2} - 3x + 9}{(x + 3)(x + 3)(x - 3)}$

Hope you could understand.

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