Bài 1.
\(\eqalign{ A &= \left( {\sqrt 6 + \sqrt {10} } \right).\sqrt {4 - \sqrt {15} } \cr& = \sqrt 2 \left( {\sqrt 3 + \sqrt 5 } \right).\sqrt {4 - \sqrt {15} } \cr & = \left( {\sqrt 3 + \sqrt 5 } \right).\sqrt {8 - 2\sqrt {15} } \cr&= \left( {\sqrt 3 + \sqrt 5 } \right).\sqrt {{{\left( {\sqrt 3 - \sqrt 5 } \right)}^2}} \cr & = \left( {\sqrt 3 + \sqrt 5 } \right).\left| {\sqrt 3 - \sqrt 5 } \right| \cr&= \left( {\sqrt 3 + \sqrt 5 } \right).\left( {\sqrt 5 - \sqrt 3 } \right)\,\,\left( {\text{Vì }\,\sqrt 3 < \sqrt 5 } \right) \cr & = {\left( {\sqrt 5 } \right)^2} - {\left( {\sqrt 3 } \right)^2} \cr&= 5 - 3 = 2 \cr} \)\(\eqalign{ B &= {{\sqrt 3 + 2} \over {\sqrt 3 - 2}} - {{\sqrt 3 - 2} \over {\sqrt 3 + 2}} + {{8\sqrt 6 - 8\sqrt 3 } \over {\sqrt 2 - 1}} \cr & = {{{{\left( {\sqrt 3 + 2} \right)}^2}} \over {\left( {\sqrt 3 - 2} \right)\left( {\sqrt 3 + 2} \right)}} - {{{{\left( {\sqrt 3 - 2} \right)}^2}} \over {\left( {\sqrt 3 - 2} \right)\left( {\sqrt 3 + 2} \right)}} + {{8\sqrt 3 \left( {\sqrt 2 - 1} \right)} \over {\sqrt 2 - 1}} \cr & = {{3 + 4\sqrt 3 + 4 - \left( {3 - 4\sqrt 3 + 4} \right)} \over {{{\left( {\sqrt 3 } \right)}^2} - {2^2}}} + 8\sqrt 3 \cr & = {{3 + 4\sqrt 3 + 4 - 3 + 4\sqrt 3 - 4} \over {3 - 4}} + 8\sqrt 3 \cr & = {{8\sqrt 3 } \over { - 1}} + 8\sqrt 3 = - 8\sqrt 3 + 8\sqrt 3 = 0 \cr} \)
Bài 2.
\(\eqalign{ Q& = \sqrt {\sqrt 2 + 2\sqrt {\sqrt 2 - 1} } + \sqrt {\sqrt 2 - 2\sqrt {\sqrt 2 - 1} } \cr & = \sqrt {{{\left( {\sqrt {\sqrt 2 - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {\sqrt 2 - 1} - 1} \right)}^2}} \cr & = \left| {\sqrt {\sqrt 2 - 1} + 1} \right| + \left| {\sqrt {\sqrt 2 - 1} - 1} \right| \cr & = \sqrt {\sqrt 2 - 1} + 1 + 1 - \sqrt {\sqrt 2 - 1} \cr&= 2\,\,\left( {\text{Vì }\,\sqrt {\sqrt 2 - 1} < 1} \right) \cr} \)
Bài 3. a. Điều kiện: \(x\ge 0\)\(\eqalign{ & \left( {2 - \sqrt x } \right)\left( {1 + \sqrt x } \right) = - x + \sqrt 5 \cr & \Leftrightarrow 2 + 2\sqrt x - \sqrt x - x = - x + \sqrt 5 \cr & \Leftrightarrow \sqrt x = \sqrt 5 - 2 \Leftrightarrow x = {\left( {\sqrt 5 - 2} \right)^2} \cr & \Leftrightarrow x = 9 - 4\sqrt 5 \ge 0\,\,\left( \text{nhận} \right) \cr} \)
b.
\(\eqalign{ & \sqrt {{x^2} + 2x\sqrt 3 + 3} = \sqrt 3 + x \cr & \Leftrightarrow \sqrt {{{\left( {x + \sqrt 3 } \right)}^2}} = \sqrt 3 + x \cr & \Leftrightarrow \left| {x + \sqrt 3 } \right| = \sqrt 3 + x \cr & \Leftrightarrow x + \sqrt 3 \ge 0 \Leftrightarrow x \ge - \sqrt 3 \cr} \)
Bài 4. a. Điều kiện để biểu thức A có nghĩa :
\(\eqalign{ & \left\{ {\matrix{ {x \ge 0} \cr {x - 1 \ge 0} \cr {\sqrt x - \sqrt {x - 1} \ne 0} \cr {1 - \sqrt x \ne 0} \cr } } \right. \Leftrightarrow x > 1 \cr & A = {1 \over {\sqrt x + \sqrt {x - 1} }} - {1 \over {\sqrt x - \sqrt {x - 1} }} - {{x\sqrt x - x} \over {1 - \sqrt x }} \cr & = {{\sqrt x - \sqrt {x - 1} } \over {\left( {\sqrt x + \sqrt {x - 1} } \right)\left( {\sqrt x - \sqrt {x - 1} } \right)}} - {{\sqrt x + \sqrt {x - 1} } \over {\left( {\sqrt x - \sqrt {x - 1} } \right)\left( {\sqrt x + \sqrt {x - 1} } \right)}} - {{x\left( {\sqrt x - 1} \right)} \over {1 - \sqrt x }} \cr & = {{\sqrt x - \sqrt {x - 1} - \left( {\sqrt x + \sqrt {x - 1} } \right)} \over {\left( {\sqrt x + \sqrt {x - 1} } \right)\left( {\sqrt x - \sqrt {x - 1} } \right)}} + {{x\left( {\sqrt x - 1} \right)} \over {\sqrt x - 1}} \cr & = {{\sqrt x - \sqrt {x - 1} - \sqrt x - \sqrt {x - 1} } \over {x - \left( {x - 1} \right)}} + x \cr & = {{ - 2\sqrt {x - 1} } \over 1} + x = - 2\sqrt {x - 1} + x \cr & = {\left( {\sqrt {x - 1} - 1} \right)^2} \cr} \)
b. \(A > 0 \Leftrightarrow \left\{ {\matrix{ {x > 1} \cr {\sqrt {x - 1} - 1 \ne 0} \cr } } \right. \Leftrightarrow \left\{ {\matrix{ {x > 1} \cr {x \ne 2} \cr } } \right.\)
Vậy để \(A > 0\) thì \(x > 1\) và \(x ≠ 2\).
