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Sagot :
Let's simplify the given expression step by step:
Given expression:
[tex]$ \frac{n+2}{2n+2} + \frac{3}{2n} $[/tex]
1. Simplify the first fraction [tex]\(\frac{n+2}{2n+2}\)[/tex]:
Notice that both the numerator and the denominator can be divided by 2:
[tex]$ \frac{n+2}{2n+2} = \frac{n+2}{2(n+1)} = \frac{n+2}{2(n+1)} $[/tex]
Thus, the simplified first fraction is:
[tex]$ \frac{n+2}{2(n+1)} $[/tex]
2. The second fraction remains as:
[tex]$ \frac{3}{2n} $[/tex]
3. To add the two fractions [tex]\(\frac{n+2}{2(n+1)} + \frac{3}{2n}\)[/tex], we need a common denominator. The common denominator for [tex]$\frac{n+2}{2(n+1)}$[/tex] and [tex]$\frac{3}{2n}$[/tex] is:
[tex]$ 2n(n+1) $[/tex]
4. Rewrite each fraction with the common denominator:
For [tex]\(\frac{n+2}{2(n+1)}\)[/tex]:
[tex]$ \frac{n+2}{2(n+1)} = \frac{(n+2)n}{2n(n+1)} = \frac{n^2 + 2n}{2n(n+1)} $[/tex]
For [tex]\(\frac{3}{2n}\)[/tex]:
[tex]$ \frac{3}{2n} = \frac{3(n+1)}{2n(n+1)} = \frac{3n + 3}{2n(n+1)} $[/tex]
5. Add the two fractions together:
[tex]$ \frac{n^2 + 2n}{2n(n+1)} + \frac{3n + 3}{2n(n+1)} = \frac{(n^2 + 2n) + (3n + 3)}{2n(n+1)} $[/tex]
6. Combine like terms in the numerator:
[tex]$ \frac{n^2 + 2n + 3n + 3}{2n(n+1)} = \frac{n^2 + 5n + 3}{2n(n+1)} $[/tex]
Hence, the simplified expression is:
[tex]$ \frac{n^2 + 5n + 3}{2n(n+1)} $[/tex]
Review the given options:
A) [tex]\(\frac{-n^2 + 3n + 3}{2n(n+1)}\)[/tex]
B) [tex]\(\frac{n - n^2 + 6}{4n(n+1)}\)[/tex]
C) [tex]\(\frac{n^2 + 5n + 3}{2n(n+1)}\)[/tex]
D) [tex]\(\frac{5}{2(2n + 1)}\)[/tex]
Comparing our result [tex]\(\frac{n^2 + 5n + 3}{2n(n+1)}\)[/tex] to the options, we see that the correct choice is:
C) [tex]\(\frac{n^2 + 5n + 3}{2n(n+1)}\)[/tex]
Given expression:
[tex]$ \frac{n+2}{2n+2} + \frac{3}{2n} $[/tex]
1. Simplify the first fraction [tex]\(\frac{n+2}{2n+2}\)[/tex]:
Notice that both the numerator and the denominator can be divided by 2:
[tex]$ \frac{n+2}{2n+2} = \frac{n+2}{2(n+1)} = \frac{n+2}{2(n+1)} $[/tex]
Thus, the simplified first fraction is:
[tex]$ \frac{n+2}{2(n+1)} $[/tex]
2. The second fraction remains as:
[tex]$ \frac{3}{2n} $[/tex]
3. To add the two fractions [tex]\(\frac{n+2}{2(n+1)} + \frac{3}{2n}\)[/tex], we need a common denominator. The common denominator for [tex]$\frac{n+2}{2(n+1)}$[/tex] and [tex]$\frac{3}{2n}$[/tex] is:
[tex]$ 2n(n+1) $[/tex]
4. Rewrite each fraction with the common denominator:
For [tex]\(\frac{n+2}{2(n+1)}\)[/tex]:
[tex]$ \frac{n+2}{2(n+1)} = \frac{(n+2)n}{2n(n+1)} = \frac{n^2 + 2n}{2n(n+1)} $[/tex]
For [tex]\(\frac{3}{2n}\)[/tex]:
[tex]$ \frac{3}{2n} = \frac{3(n+1)}{2n(n+1)} = \frac{3n + 3}{2n(n+1)} $[/tex]
5. Add the two fractions together:
[tex]$ \frac{n^2 + 2n}{2n(n+1)} + \frac{3n + 3}{2n(n+1)} = \frac{(n^2 + 2n) + (3n + 3)}{2n(n+1)} $[/tex]
6. Combine like terms in the numerator:
[tex]$ \frac{n^2 + 2n + 3n + 3}{2n(n+1)} = \frac{n^2 + 5n + 3}{2n(n+1)} $[/tex]
Hence, the simplified expression is:
[tex]$ \frac{n^2 + 5n + 3}{2n(n+1)} $[/tex]
Review the given options:
A) [tex]\(\frac{-n^2 + 3n + 3}{2n(n+1)}\)[/tex]
B) [tex]\(\frac{n - n^2 + 6}{4n(n+1)}\)[/tex]
C) [tex]\(\frac{n^2 + 5n + 3}{2n(n+1)}\)[/tex]
D) [tex]\(\frac{5}{2(2n + 1)}\)[/tex]
Comparing our result [tex]\(\frac{n^2 + 5n + 3}{2n(n+1)}\)[/tex] to the options, we see that the correct choice is:
C) [tex]\(\frac{n^2 + 5n + 3}{2n(n+1)}\)[/tex]
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