Respuesta :

caylus
Hello,

As you write 1/x, x must be different of 0.
[tex] \dfrac{1}{x} = \dfrac{x+3}{2x^2} \\ ==\textgreater\ 2x^2=x^2+3x \cross`\product \\ ==\textgreater\ x^2-3x =0\\ ==\textgreater\ x(x-3)=0 \\ ==\textgreater\ x=0 (to exclude) or x=3\\ Answer\ x=3 [/tex]


Answer:

D) x=3

Step-by-step explanation:

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