A) \[{{(a+b)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)\]
B) \[{{(a-b)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}={{a}^{3}}-{{b}^{3}}-3ab(a-b)\]
C) \[{{a}^{3}}+{{b}^{3}}=(a+b)({{a}^{2}}-ab+{{b}^{2}})\]
D) none of these
Correct Answer: A
Solution :
\[x+\frac{1}{1+\frac{1}{1+\frac{1}{x}}}=\frac{1}{2}+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}\] \[=\frac{1}{2}+\frac{1}{1+\frac{1}{1+2}}=\frac{1}{2}+\frac{1}{1+\frac{1}{3}}=\frac{1}{2}+\frac{1}{\frac{4}{3}}\] \[=\frac{1}{2}+\frac{3}{4}=\frac{2+3}{4}=\frac{5}{4}\]
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