Question

Hi ... i appreciate you to answer this questuion...

$\frac{ {x}^{2} }{ {x}^{4} + 1 }$
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Answer 1

Answer:

$\large \boxed{\sf \ \ \dfrac{x^2}{x^4+1}=\dfrac{1}{7} \ \ }$

Step-by-step explanation:

Hello,

We know that (let's assume that x is different from 0 as we cannot divide by 0)

   $x+\dfrac{1}{x}=3$

and we want to estimate

   $\dfrac{x^2}{x^4+1}$

Let's take the square.

$9=3^2=(x+\dfrac{1}{x})^2=x^2+2\cdot x \cdot \dfrac{1}{x}+\dfrac{1}{x^2}=x^2+2+\dfrac{1}{x^2}=\dfrac{x^4+1}{x^2}+2$

So, we can write

$\dfrac{x^4+1}{x^2}=9-2=7 \\ \\\\\text{*** let's take the inverse ***} \\ \\\dfrac{x^2}{x^4+1}=\dfrac{1}{7}$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you