Bài 1.
a. A có nghĩa \( \Leftrightarrow {{ - 3} \over {3 - x}} \ge 0 \Leftrightarrow 3 - x < 0 \Leftrightarrow x > 3\)
b. B có nghĩa \( \Leftrightarrow x + {1 \over x} \ge 0 \Leftrightarrow {{{x^2} + 1} \over x} \ge 0 \Leftrightarrow x > 0\)
(vì \({x^2} \ge 0,\) với mọi \(x ∈ \mathbb R\) nên \({x^2} + 1 \ge 1 > 0,\) với mọi \(x ∈\mathbb R\)).
Bài 2. a. Ta có:
\(\eqalign{ M &= {\left( {\sqrt 2 } \right)^2} - \sqrt 2 .\sqrt {3 - \sqrt 5 } + \sqrt {20} \cr & = 2 - \sqrt {6 - 2\sqrt 5 } + \sqrt {20} \cr & = 2 - \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} + \sqrt {4.5} \cr & = 2 - \left( {\sqrt 5 - 1} \right) + 2\sqrt 5 = 3 + \sqrt 5 \cr} \)
b. Ta có:
\(\eqalign{ N &= \left( {{{\sqrt 2 \left( {\sqrt 3 - 1} \right)} \over {1 - \sqrt 3 }} - \sqrt 5 } \right)\left( {\sqrt 5 - \sqrt 2 } \right) \cr & = - \left( {\sqrt 2 + \sqrt 5 } \right)\left( {\sqrt 5 - \sqrt 2 } \right) \cr & = - \left( {5 - 2} \right) = - 3 \cr} \)
Bài 3. a. Ta có:
\(\eqalign{ P &= {{a\sqrt a } \over {\sqrt a - 1}} - {1 \over {\sqrt a - 1}} \cr&= {{{{\left( {\sqrt a } \right)}^3} - {1^3}} \over {\sqrt a - 1}} \cr & = {{\left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)} \over {\sqrt a - 1}} \cr & = a + \sqrt a + 1 \cr} \)
b. Khi \(a = {9 \over 4} \Rightarrow P = {9 \over 4} + \sqrt {{9 \over 4}} + 1 = {{19} \over 4}\)
Bài 4. a. Ta có:
\(\eqalign{ & \sqrt {4{x^2} - 4x + 1} = 3\cr& \Leftrightarrow \sqrt {{{\left( {2x - 1} \right)}^2}} = 3 \cr & \Leftrightarrow \left| {2x - 1} \right| = 3\cr& \Leftrightarrow \left[ {\matrix{ {2x - 1 = 3} \cr {2x - 1 = - 3} \cr } } \right. \Leftrightarrow \left[ {\matrix{ {x = 2} \cr {x = - 1} \cr } } \right. \cr} \)
b. Ta có:
\(\eqalign{ & 3\left( {\sqrt x + 2} \right) + 5 = 4\sqrt {4x} + 1 \cr & \Leftrightarrow 3\sqrt x + 6 + 5 = 8\sqrt x + 1 \cr & \Leftrightarrow 5\sqrt x = 10 \Leftrightarrow \sqrt x = 2 \cr & \Leftrightarrow x = 4 \cr} \)
Bài 5. Ta có:
\(\eqalign{ & \sqrt {1 - 3x} < 2 \Leftrightarrow \left\{ {\matrix{ {1 - 3x \ge 0} \cr {1 - 3x < 4} \cr } } \right. \cr & \Leftrightarrow \left\{ {\matrix{ {x \le {1 \over 3}} \cr {x > - 1} \cr } } \right. \Leftrightarrow - 1 < x \le {1 \over 3} \cr} \)
