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Question-89172




Question Number 89172 by M±th+et£s last updated on 15/Apr/20
Answered by mahdi last updated on 15/Apr/20
(((x^1 .x^((−b)/a) .x^(1/a) .x^(b/a^2 ) )/(x^(b/a^2 ) .x^(a/a) )))^(1/(b−1)) =(x^((1−b)/a) )^(1/(b−1)) =x^((−1)/a) =(1/( (x)^(1/a) ))
$$\left(\frac{\mathrm{x}^{\mathrm{1}} .\mathrm{x}^{\frac{−\mathrm{b}}{\mathrm{a}}} .\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{a}}} .\mathrm{x}^{\frac{\mathrm{b}}{\mathrm{a}^{\mathrm{2}} }} }{\mathrm{x}^{\frac{\mathrm{b}}{\mathrm{a}^{\mathrm{2}} }} .\mathrm{x}^{\frac{\mathrm{a}}{\mathrm{a}}} }\right)^{\frac{\mathrm{1}}{\mathrm{b}−\mathrm{1}}} =\left(\mathrm{x}^{\frac{\mathrm{1}−\mathrm{b}}{\mathrm{a}}} \right)^{\frac{\mathrm{1}}{\mathrm{b}−\mathrm{1}}} =\mathrm{x}^{\frac{−\mathrm{1}}{\mathrm{a}}} =\frac{\mathrm{1}}{\:\sqrt[{\mathrm{a}}]{\mathrm{x}}} \\ $$
Commented by M±th+et£s last updated on 15/Apr/20
perfect. thank you sir
$${perfect}.\:{thank}\:{you}\:{sir} \\ $$

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