htx22
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  • 25-04-2017
  • Mathematics
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[tex] x^{2} \sqrt{x} \sqrt{x} \sqrt[n]{x} x_{123} [/tex]

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cause
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  • 25-04-2017
x^2\sqrt[n]{x}\sqrt{x}\sqrt{x}=\:x^{2+\frac{1}{2}+\frac{1}{2}+\frac{1}{n}}=\:x^{\frac{1}{n}+3} =x^{\frac{1}{n}+3}x_{123}
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