Respuesta :

Given:

[tex]\sqrt{15(}x-1)\div\sqrt{2x^2}[/tex]

Required :

To calculate which option is correct

Explanation:

[tex]\begin{gathered} in\text{ the range of real numbers, the even root sign cannot be negative} \\ \\ so\text{ 15\lparen x-1\rparen}\ge0 \\ \\ so\text{ }x\ge1 \\ \\ And\text{ 2x}^2\ge0\text{ is established} \\ \end{gathered}[/tex]

Required answer:

Option B

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