Question

Help!!
Simplify:

[((√(x^2+1)-x) x (2x/(2√(x^2+1))]/(x^2+1)

Answer 1

\[(\sqrt{x^2+1}-x \times (2x/2\sqrt{x^2+1}))/\sqrt{x^2+1}\]

Answer 2

Is this it?

\(\dfrac{(\sqrt{x^2 + 1} - x)\frac{2x}{2 \sqrt{x^2 + 1} }}{x^2 + 1} \)

Answer 3

Yeah

Answer 4

\(=\dfrac{(\sqrt{x^2 + 1} - x)\frac{\cancel{2}x}{\cancel{2} \sqrt{x^2 + 1} + 1}}{x^2 + 1} \)

\(= \dfrac{(\sqrt{x^2 + 1} - x)\frac{x}{ \sqrt{x^2 + 1} + 1}}{x^2 + 1} \)

\(= \dfrac{x(\sqrt{x^2 + 1} - x) }{(x^2 + 1)( \sqrt{x^2 + 1} + 1 )} \)

Answer 5

where did the 1+1 come from??

Answer 6

You're right. I made a mistake when I copied one step to the next.

Answer 7

\(=\dfrac{(\sqrt{x^2 + 1} - x)\frac{\cancel{2}x}{\cancel{2} \sqrt{x^2 + 1}}}{x^2 + 1}\)

\(=\dfrac{(\sqrt{x^2 + 1} - x)\frac{x}{\sqrt{x^2 + 1}}}{x^2 + 1}\)

\(=\dfrac{x(\sqrt{x^2 + 1} - x)} {(x^2 + 1) \sqrt{x^2 + 1} }\)

Now we can rationalize the denominator:

\(=\dfrac{x(\sqrt{x^2 + 1} - x)} {(x^2 + 1) \sqrt{x^2 + 1} } \times \dfrac{ \sqrt{x^2 + 1} }{ \sqrt{x^2 + 1} }\)

\(=\dfrac{x \sqrt{x^2 + 1} (\sqrt{x^2 + 1} - x)} {(x^2 + 1) ^2 }\)