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mariarad [96]
3 years ago
7

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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