A) \[\frac{1}{2}\]
B) \[2\]
C) \[\frac{1}{20}\]
D) \[20\]
Correct Answer: A
Solution :
[a] If \[a=11\]then \[b=9,\] \[\frac{{{a}^{2}}+{{b}^{2}}+ab}{{{a}^{3}}-{{b}^{3}}}=\frac{{{a}^{2}}+{{b}^{2}}+ab}{(a-b)({{a}^{2}}+{{b}^{2}}+ab)}\] \[=\frac{1}{(a-b)}=\frac{1}{11-9}=\frac{1}{2}\] |
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