3 years ago

Find the sum. 5 2a a? + 2a + 1 + a2 + 4a + 3

Mathematics

2 answers:

Vinvika [58]3 years ago

7 0

Answer:

$= \frac{2 {a}^{2} + 11a + 5}{ {(a + 1)}^{2}(a + 3)} \\$

Step-by-step explanation:

$\frac{2a}{ {a}^{2} + 2a + 1} + \frac{5}{ {a}^{2} + 4a + 3 } \\ \frac{2a}{(a + 1)(a + 1)} + \frac{5}{(a + 1)(a + 3)} \\ \frac{2a}{ {(a + 1)}^{2} } + \frac{5}{(a + 1)(a + 3)} \\ \frac{2a(a + 3) + 5(a + 1)}{ {(a + 1)}^{2}(a + 3) } \\ \frac{2 {a}^{2} + 6a + 5a + 5 }{ {(a + 1)}^{2}(a + 3)} \\ = \frac{2 {a}^{2} + 11a + 5}{ {(a + 1)}^{2}(a + 3)}$

Salsk061 [2.6K]3 years ago

5 0

$\frac{2a}{a^2+2a+1}+\frac{5}{a^2+4a+3}$

Factor the denominators.

$\frac{2a}{\left(a+1\right)^2}+\frac{5}{\left(a+1\right)\left(a+3\right)}$

Adjust fractions based on LCM.

$\frac{2a\left(a+3\right)}{\left(a+1\right)^2\left(a+3\right)}+\frac{5\left(a+1\right)}{\left(a+1\right)^2\left(a+3\right)}$

Denominators are same, so add the fractions.

$\frac{2a\left(a+3\right)+5\left(a+1\right)}{\left(a+1\right)^2\left(a+3\right)}$

Expand the numerator.

$\frac{2a^2+11a+5}{\left(a+1\right)^2\left(a+3\right)}$

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