Calculation:

The denominator of a rational expression can be rationalized by removing the irrational term from it.

To remove the irration term, the numerator and denominator of the expression are multiplied by the conjugate of the denominator.

Conjugate of an expression \(\displaystyle\sqrt{{{a}}}+\sqrt{{{b}}}\) is given as \(\displaystyle\sqrt{{{a}}}-\sqrt{{{b}}}\) and vice-versa.

So the conjugate of the denominator is given as \(\displaystyle\sqrt{{{2}}}-{x}\)

Multiplying the numerator and the denominator with the conjugate,

\(\displaystyle{\frac{{{1}}}{{\sqrt{{{2}}}+{x}}}}={\frac{{{1}}}{{\sqrt{{{2}}}+{x}}}}\times{\frac{{\sqrt{{{2}}}-{x}}}{{\sqrt{{{2}}}-{x}}}}\)

\(\displaystyle{\frac{{{1}}}{{\sqrt{{{2}}}+{x}}}}={\frac{{\sqrt{{{2}}}-{x}}}{{{\left(\sqrt{{{2}}}+{x}\right)}{\left(\sqrt{{{2}}}-{x}\right)}}}}\)

\(\displaystyle{\frac{{{1}}}{{\sqrt{{{2}}}+{x}}}}={\frac{{\sqrt{{{2}}}-{x}}}{{{\left(\sqrt{{{2}}}\right)}^{{{2}}}-{\left({x}\right)}^{{{2}}}}}}{i}.{e}.{\left({a}+{b}\right)}{\left({a}-{c}\right)}={a}^{{{2}}}-{b}^{{{2}}}\)

\(\displaystyle{\frac{{{1}}}{{\sqrt{{{2}}}+{x}}}}={\frac{{\sqrt{{{2}}}-{x}}}{{{2}-{x}^{{{2}}}}}}\) (Since square and square root are opposite operations, they cancel out).

The denominator of a rational expression can be rationalized by removing the irrational term from it.

To remove the irration term, the numerator and denominator of the expression are multiplied by the conjugate of the denominator.

Conjugate of an expression \(\displaystyle\sqrt{{{a}}}+\sqrt{{{b}}}\) is given as \(\displaystyle\sqrt{{{a}}}-\sqrt{{{b}}}\) and vice-versa.

So the conjugate of the denominator is given as \(\displaystyle\sqrt{{{2}}}-{x}\)

Multiplying the numerator and the denominator with the conjugate,

\(\displaystyle{\frac{{{1}}}{{\sqrt{{{2}}}+{x}}}}={\frac{{{1}}}{{\sqrt{{{2}}}+{x}}}}\times{\frac{{\sqrt{{{2}}}-{x}}}{{\sqrt{{{2}}}-{x}}}}\)

\(\displaystyle{\frac{{{1}}}{{\sqrt{{{2}}}+{x}}}}={\frac{{\sqrt{{{2}}}-{x}}}{{{\left(\sqrt{{{2}}}+{x}\right)}{\left(\sqrt{{{2}}}-{x}\right)}}}}\)

\(\displaystyle{\frac{{{1}}}{{\sqrt{{{2}}}+{x}}}}={\frac{{\sqrt{{{2}}}-{x}}}{{{\left(\sqrt{{{2}}}\right)}^{{{2}}}-{\left({x}\right)}^{{{2}}}}}}{i}.{e}.{\left({a}+{b}\right)}{\left({a}-{c}\right)}={a}^{{{2}}}-{b}^{{{2}}}\)

\(\displaystyle{\frac{{{1}}}{{\sqrt{{{2}}}+{x}}}}={\frac{{\sqrt{{{2}}}-{x}}}{{{2}-{x}^{{{2}}}}}}\) (Since square and square root are opposite operations, they cancel out).