Given,

√2 = 1.4142

Given EXPRESSION is,

$(\BEGIN{array}{l}2\sqrt 2+ \sqrt 2+ \frac{1}{{2 + \surd 2}} + \frac{1}{{\sqrt 2- 2}}\\ = 2\sqrt 2+ \sqrt 2+ \frac{{\left( {2 - \sqrt 2 } \right)}}{{\left( {2 + \sqrt 2 } \right)\left( {2 - \sqrt 2 } \right)}} + \frac{{\left( {\sqrt 2+ 2} \right)}}{{\left( {\sqrt 2- 2} \right)\left( {\sqrt 2+ 2} \right)}}\end{array})$

USING the identity: a² – b² = (a + b)(a – b)

$(\begin{array}{l} = 2\sqrt 2+ \sqrt 2+ \frac{{\left( {2 - \sqrt 2 } \right)}}{{{2^2} - {{\left( {\sqrt 2 } \right)}^2}}} + \frac{{\left( {\sqrt 2+ 2} \right)}}{{{{\left( {\sqrt 2 } \right)}^2} - {2^2}}}\\ = 2\sqrt 2+ \sqrt 2+ \frac{{\left( {2 - \sqrt 2 } \right)}}{{4 - 2}} + \frac{{\left( {\sqrt 2+ 2} \right)}}{{2 - 4}}\\ = 2\sqrt 2+ \sqrt 2+ \frac{{\left( {2 - \sqrt 2 } \right)}}{2} - \frac{{\left( {\sqrt 2+ 2} \right)}}{2}\\ = 2\sqrt 2+ \sqrt 2+ \frac{{2 - \sqrt 2- \sqrt 2- 2}}{2}\\ = 2\sqrt 2+ \sqrt 2- \frac{{2\sqrt 2 }}{2}\end{array})$

= 2√2+√2- √2

= 2√2

Putting value of √2 = 1.4142