Bài 1. a.
\(\eqalign{ & \sqrt {7 - 4\sqrt 3 } - \sqrt {4 + 2\sqrt 3 } \cr&= \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {1 + \sqrt 3 } \right)}^2}} \cr & = \left| {2 - \sqrt 3 } \right| - \left( {1 + \sqrt 3 } \right) \cr&= 2 - \sqrt 3 - 1 - \sqrt 3 \,\,\left( {vi\,2 > \sqrt 3 } \right) \cr & = 1 - 2\sqrt 3 \cr} \)
b.
\(\eqalign{ & \left( {{{14} \over {\sqrt {14} }} + {{\sqrt {12} + \sqrt {30} } \over {\sqrt 2 + \sqrt 5 }}} \right).\sqrt {5 - \sqrt {21} } \cr & = \left[ {\sqrt {14} + {{\sqrt 6 \left( {\sqrt 2 + \sqrt 5 } \right)} \over {\sqrt 2 + \sqrt 5 }}} \right].\sqrt {5 - \sqrt {21} } \cr & = \left( {\sqrt {14} + \sqrt 6 } \right).\sqrt {5 - \sqrt {21} } \cr & = \sqrt 2 \left( {\sqrt 7 + \sqrt 3 } \right).\sqrt {5 - \sqrt {21} } \cr & = \left( {\sqrt 7 + \sqrt 3 } \right).\sqrt {10 - 2\sqrt {21} } \cr & = \left( {\sqrt 7 + \sqrt 3 } \right).\sqrt {{{\left( {\sqrt 7 - \sqrt 3 } \right)}^2}} \cr & = \left( {\sqrt 7 + \sqrt 3 } \right).\left| {\sqrt 7 - \sqrt 3 } \right| \cr & = \left( {\sqrt 7 + \sqrt 3 } \right).\left( {\sqrt 7 - \sqrt 3 } \right)\,\,\left( {\text{ vì }\,\sqrt 7 > \sqrt 3 } \right) \cr & = {\left( {\sqrt 7 } \right)^2} - {\left( {\sqrt 3 } \right)^2} = 4. \cr} \)
Bài 2. Biến đổi vế trái ta có:
\(\eqalign{ & {4 \over {\sqrt x + 2}} + {2 \over {\sqrt x - 2}} - {{5\sqrt x - 6} \over {\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} \cr & = {{4\left( {\sqrt x - 2} \right) + 2\left( {\sqrt x + 2} \right) - \left( {5\sqrt x - 6} \right)} \over {\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}} \cr & = {{4\sqrt x - 8 + 2\sqrt x + 4 - 5\sqrt x + 6} \over {\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}} \cr & = {{\sqrt x + 2} \over {\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}} = {1 \over {\sqrt x - 2}} \cr} \)
Bài 3. a.
\(\eqalign{ & P = \left( {{1 \over {x - \sqrt x }} + {{\sqrt x } \over {x - 1}}} \right):{{x\sqrt x - 1} \over {x\sqrt x - \sqrt x }} \cr & = \left[ {{1 \over {\sqrt x \left( {\sqrt x - 1} \right)}} + {{\sqrt x } \over {\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right]:{{\sqrt {{x^3}} - 1} \over {\sqrt x \left( {x - 1} \right)}} \cr & = {{\sqrt x + 1 + x} \over {\sqrt x \left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.{{\sqrt x \left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)} \over {\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} \cr&= {1 \over {\sqrt x - 1}} \cr} \)
(với \(x > 0\) và \(x ≠ 1\))
b. \(P = {1 \over 2} \Leftrightarrow {1 \over {\sqrt x - 1}} = {1 \over 2} \Leftrightarrow \sqrt x - 1 = 2 \)
\(\Leftrightarrow \sqrt x = 3\) \(⇔ x = 9\) (thỏa mãn điều kiện \(x > 0\) và \(x ≠ 1\))
