## giải giúp mình với pleaseee

Question

$$\begin{array}{l} 1)A = \left( {1 – \dfrac{{\sqrt 5 \left( {\sqrt 5 + 1} \right)}}{{\sqrt 5 + 1}}} \right).\left( {\dfrac{{\sqrt 5 \left( {\sqrt 5 – 1} \right)}}{{1 – \sqrt 5 }} – 1} \right)\\ = \left( {1 – \sqrt 5 } \right)\left( { – \dfrac{{\sqrt 5 \left( {\sqrt 5 – 1} \right)}}{{\sqrt 5 – 1}} – 1} \right)\\ = – \left( {1 – \sqrt 5 } \right)\left( {1 + \sqrt 5 } \right)\\ = – \left( {1 – 5} \right) = 4\\ 2)B = \left[ {\dfrac{{3.5\sqrt 5 }}{{15}} – \dfrac{{\left( {10 – 4\sqrt 6 } \right)\left( {\sqrt 5 + 2} \right)}}{{5 – 4}}} \right].\dfrac{1}{{\sqrt 5 }}\\ = \left( {\sqrt 5 – \left( {10 – 4\sqrt 6 } \right)\left( {\sqrt 5 + 2} \right)} \right).\dfrac{1}{{\sqrt 5 }}\\ = \left( {\sqrt 5 – 10\sqrt 5 – 20 + 4\sqrt {30} + 8\sqrt 6 } \right).\dfrac{1}{{\sqrt 5 }}\\ = \left( { – 9\sqrt 5 – 20 + 4\sqrt {30} + 8\sqrt 6 } \right).\dfrac{1}{{\sqrt 5 }}\\ = – 9 – 4\sqrt 5 + 4\sqrt 6 + \dfrac{{8\sqrt 6 }}{{\sqrt 5 }}\\ 3)C = \left[ {\dfrac{{10\sqrt {10} }}{{100}} – \dfrac{{\sqrt {10} \left( {\sqrt 5 – \sqrt 2 } \right)}}{{2\left( {\sqrt 5 – \sqrt 2 } \right)}}} \right].\sqrt {10} \\ = \left[ {\dfrac{{\sqrt {10} }}{{10}} – \dfrac{{\sqrt {10} }}{2}} \right].\sqrt {10} \\ = 1 – 5 = – 4\\ 4)D = \dfrac{1}{{\sqrt {25 + 2.2\sqrt 6 .5 + 24} }} – \dfrac{1}{{\sqrt {25 – 2.2\sqrt 6 .5 + 24} }} + \dfrac{1}{{\sqrt {4 – 2.2.\sqrt 3 + 3} }}\\ = \dfrac{1}{{\sqrt {{{\left( {5 + 2\sqrt 6 } \right)}^2}} }} – \dfrac{1}{{\sqrt {{{\left( {5 – 2\sqrt 6 } \right)}^2}} }} + \dfrac{1}{{\sqrt {{{\left( {2 – \sqrt 3 } \right)}^2}} }}\\ = \dfrac{1}{{5 + 2\sqrt 6 }} – \dfrac{1}{{5 – 2\sqrt 6 }} + \dfrac{1}{{2 – \sqrt 3 }}\\ = \dfrac{{5 – 2\sqrt 6 – 5 – 2\sqrt 6 }}{{25 – 24}} + \dfrac{{2 + \sqrt 3 }}{{4 – 3}}\\ = – 4\sqrt 6 + 2 + \sqrt 3 \end{array}$$