A) \[\frac{1}{2}\]
B) 2
C) \[\frac{1}{20}\]
D) 20
Correct Answer: A
Solution :
[a] \[\frac{{{a}^{2}}+{{b}^{2}}+ab}{{{a}^{3}}-\,{{b}^{3}}}\] \[=\frac{{{a}^{2}}+{{b}^{2}}=ab}{(a-b)\,\,({{a}^{2}}+{{b}^{2}}+ab)}=\frac{1}{a-b}=\frac{1}{11-9}=\frac{1}{2}\]You need to login to perform this action.
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