A) \[{{\text{a}}^{\text{2}}}+\text{a}+\text{1}\]
B) \[{{\text{a}}^{\text{2}}}-\text{a}+\text{1}\]
C) \[{{\text{a}}^{\text{2}}}-\text{a}-\text{1}\]
D) \[{{\text{a}}^{\text{2}}}+\text{a}-\text{1}\]
Correct Answer: B
Solution :
Let\[\text{216 }=\text{ 192 }\times \text{ 1 }+\text{ 24}\] \[\text{192 }=\text{ 24 }\times \text{ 8 }+\text{ }0\]and\[{{6}^{x}}\] Then\[{{6}^{x}}={{(2\times 3)}^{x}}={{2}^{x}}\times {{3}^{x}}\] \[\Rightarrow \]\[{{6}^{x}}\] Hence, \[\therefore \]
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