A) 2a
B) \[x+y+z\]
C) 1
D) 0
Correct Answer: C
Solution :
(c): \[\frac{{{\left( {{a}^{x+y}} \right)}^{2}}{{\left( {{a}^{y+z}} \right)}^{2}}{{\left( {{a}^{z+x}} \right)}^{2}}}{\left( {{a}^{4x}}.{{a}^{4y}}.{{a}^{4z}} \right)}\Rightarrow \frac{{{a}^{2x+2y}}.{{a}^{2y+2z}}.{{a}^{2z+2x}}}{{{a}^{4x}}.{{a}^{4y}}.{{a}^{4z}}}\] \[\Rightarrow \frac{{{a}^{4x+4y+4z}}}{{{a}^{4x+4y+4z}}}=1\]You need to login to perform this action.
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