$\begin{array}{l} a)\sqrt {8 - 2\sqrt 7 } = \sqrt {{{\left( {\sqrt 7 - 1} \right)}^2}} = \sqrt 7 - 1\\ b){\left( {\sqrt {2 - \sqrt 3 } - \sqrt {2 + \sqrt 3 } } \right)^2} = 2 - \sqrt 3 + 2 + \sqrt 3 - 2\sqrt {4 - 3} = 4 - 2 = 2\\ c)\left( {1 + \sqrt 3 + \sqrt 5 } \right)\left( {1 + \sqrt 3 - \sqrt 5 } \right)\\ = {\left( {1 + \sqrt 3 } \right)^2} - 5 = 3 + 1 + 2\sqrt 3 - 5 = 2\sqrt 3 - 1\\ d)\left( {1 + \sqrt 2 + \sqrt 3 } \right)\left( {1 - \sqrt 2 - \sqrt 3 } \right)\\ = 1 - {\left( {\sqrt 2 + \sqrt 3 } \right)^2} = 1 - \left( {2 + 3 + 2\sqrt 6 } \right) = - 4 + 2\sqrt 6 \\ e)A = \left( {\sqrt {2 + \sqrt 3 } - \sqrt {2 - \sqrt 3 } } \right):\sqrt 2 \\ \Rightarrow 2A = \sqrt {4 + 2\sqrt 3 } - \sqrt {4 - 2\sqrt 3 } = \sqrt 3 + 1 - \left( {\sqrt 3 - 1} \right) = 2\\ \Rightarrow A = 1\\ f)\left( {\sqrt {10} - \sqrt 6 } \right)\sqrt {4 - \sqrt {15} } = \sqrt 2 .\sqrt {4 - \sqrt {15} } \left( {\sqrt 5 - \sqrt 3 } \right)\\ = \sqrt {8 - 2\sqrt {15} } \left( {\sqrt 5 - \sqrt 2 } \right) = \sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} \left( {\sqrt 5 - \sqrt 3 } \right)\\ = {\left( {\sqrt 5 - \sqrt 3 } \right)^2} = 5 + 3 - 2\sqrt {15} = 8 - 2\sqrt {15} \\ 3)\\ A = \sqrt {\dfrac{{{{\left( {x - 5} \right)}^4}}}{{{{\left( {4 - x} \right)}^2}}}} - \dfrac{{{x^2} - 25}}{{x - 4}}\\ A = \dfrac{{{{\left( {x - 5} \right)}^2}}}{{\left| {4 - x} \right|}} - \dfrac{{{x^2} - 25}}{{x - 4}}\\ A = \dfrac{{{x^2} - 10x + 25}}{{4 - x}} + \dfrac{{{x^2} - 25}}{{4 - x}}\\ A = \dfrac{{2{x^2} - 10x}}{{4 - x}} = \dfrac{{2x\left( {x - 5} \right)}}{{4 - x}}\\ B = \sqrt {x - 2 + 2\sqrt {x - 3} } - \sqrt {x - 3} \\ B = \sqrt {x - 3 + 1 + 2\sqrt {x - 3} } - \sqrt {x - 3} \\ B = \sqrt {{{\left( {\sqrt {x - 3} + 1} \right)}^2}} - \sqrt {x - 3} \\ B = \sqrt {x - 3} + 1 - \sqrt {x - 3} = 1 \end{array}$
