$$\eqalign{
& a)\,\,A = {{3 - \sqrt {12} } \over {\sqrt 3 - 2}} - {1 \over {\sqrt 3 - 2}} - \sqrt {7 - 4\sqrt 3 } \cr
& A = {{3 - 2\sqrt 3 } \over {\sqrt 3 - 2}} - {{\sqrt 3 + 2} \over {\left( {\sqrt 3 - 2} \right)\left( {\sqrt 3 + 2} \right)}} - \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} \cr
& A = {{\sqrt 3 \left( {\sqrt 3 - 2} \right)} \over {\sqrt 3 - 2}} - {{\sqrt 3 + 2} \over { - 1}} - \left| {2 - \sqrt 3 } \right| \cr
& A = \sqrt 3 + \sqrt 3 + 2 - 2 + \sqrt 3 = 3\sqrt 3 \cr
& b)\,\,B = {1 \over {\sqrt x - 1}} - {1 \over {\sqrt x + 1}} + 2\,\,\left( {x \ge 0;x \ne 1} \right) \cr
& B = {{\sqrt x + 1 - \sqrt x + 1 + 2\left( {x - 1} \right)} \over {x - 1}} \cr
& B = {{2 + 2x - 2} \over {x - 1}} = {{2x} \over {x - 1}} \cr
& B = {A \over {\sqrt 3 }} = {{3\sqrt 3 } \over {\sqrt 3 }} = 3 \cr
& \Leftrightarrow {{2x} \over {x - 1}} = 3 \Leftrightarrow 2x = 3x - 3 \cr
& \Leftrightarrow x = 3\,\,\left( {tm} \right) \cr} $$
