A) 1
B) 0
C) -1
D) 10
E) None of these
Correct Answer: A
Solution :
Explanation \[=\,\,\,\frac{{{2}^{x+3}}\times {{3}^{2x-y}}\times {{5}^{x+y+3}}\times \,\,{{6}^{y\,+\,1}}}{{{6}^{x\,+\,1}}\times {{10}^{y\,+\,3}}\times {{15}^{x}}}\] \[=\,\,\,\frac{{{2}^{x+3}}\times {{3}^{2x-y}}\times {{5}^{x+y+3}}\times \,\,{{(2\times 3)}^{y\,+\,1}}}{{{(2\times 3)}^{x\,+\,1}}\times {{(5\times 2)}^{y\,+\,3}}\times {{(5\times 3)}^{x}}}\] \[=\,\,\,\frac{{{2}^{x+3}}\times {{3}^{2x-y}}\times {{5}^{x+y+3}}\times \,\,{{2}^{y\,+\,1}}\times {{3}^{y\,+\,1}}}{{{2}^{x\,+\,1}}\times {{3}^{x\,+\,1}}\times {{5}^{y\,+\,3}}\times {{2}^{y\,+\,3}}\times {{5}^{x}}\times {{3}^{x}}}\] \[=\,\,\frac{{{2}^{x+y+4}}\times {{3}^{2x\,+\,1}}\times {{5}^{x\,+\,y\,+\,3}}}{{{2}^{x+y+4}}\times {{3}^{2x+1}}\times {{5}^{x+y+3}}}=1\]You need to login to perform this action.
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