Answer:
[tex]\frac{6x-14}{x^{4} +4x}[/tex]
Step-by-step explanation:
I have to [tex]\frac{\frac{3x-7}{x^{2} } }{\frac{x^{2} }{2}+\frac{2}{x}}[/tex]
Let's start by joining the macro denominator with a common denominator. So, by applying a minimum common multiple [tex]\frac{x^{2} }{2} +\frac{2}{x}=\frac{x^{3}+ 4 }{2x}[/tex]
Now I can write the expression as
[tex]\frac{\frac{3x-7}{x^{2}}}{\frac{x^{3}+ 4 }{2x}}[/tex]
Now to convert both fractions into one, I multiply the numerator of the one above by the denominator of the one below, and the denominator of the one above with the numerator below, remaining that way.
[tex]\frac{\frac{3x-7}{x^{2}}}{\frac{x^{3}+4}{2x}}=\frac{(3x-7)(2x)}{(x^{2})(x^{3}+ 4)}[/tex]
Having the fraction in this way, I could simplify the x of the "2x" of the numerator with an x^2 (x^2=x*x) of the denominator
[tex]\frac{(3x-7)(2x)}{(x^{2})(x^{3}+4)}=\frac{2(3x-7)}{x(x^{3}+ 4)}[/tex]
finally, applying distributive property, I have to
[tex]\frac{(6x-14)}{(x^{4}+ 4x)}[/tex]
Done
