Question

Please answer both..... it's urgent ​

Answer 1

Answer:

1/abx is the answer for the given problem

Answer 2

Answer:

$\frac{ {x}^{2} \times {y}^{3} \times mn }{ {x}^{11} {y}^{12} {m}^{2} {n}^{2} } \times \frac{ {x}^{8} {y}^{9}mn }{ab} \\ = \frac{ {x}^{2} \times {y}^{3} \times mn \times {x}^{8} {y}^{9} mn }{ {x}^{11} {y}^{12} {m}^{2} {n}^{2} \times ab} \\ = \frac{ {x}^{2} \times {x}^{8} \times {y}^{3} \times {y}^{9} \times {m}^{1} {n}^{1} \times {m}^{1} {n}^{1} }{ {x}^{11} \times {y}^{12} \times {m}^{2} \times {n}^{2} \times ab } \\ = \frac{ {x}^{2 + 8} \times {y}^{3 + 9} \times {m}^{1 + 1} \times {n}^{1 + 1} }{ {x}^{11} \times {y}^{12} \times {m}^{2} \times {n}^{2} \times ab } \\ = \frac{ {x}^{10} \times {y}^{12} \times {m}^{2} \times {n}^{2} }{ {x}^{11} \times {y}^{12} \times {m}^{2} \times {n}^{2} \times ab } \\ = \frac{ {x}^{10 - 11} \times {y}^{12 - 12} \times {m}^{2 - 2} \times {n}^{2 - 2} }{ab} \\ = \frac{ {x}^{ - 1} \times {y}^{0} \times {m}^{0} \times {n}^{0} }{ab} \\ = \frac{ {x}^{ - 1} \times 1 \times 1 \times 1}{ab} \\ = \frac{ {x}^{ - 1} }{ab} \\ = \frac{1}{x \times ab}$

I hope it will help you