Giải phương trình: a, √(2x - 1) + √(x - 2) = √(x + 1) b, √(3x + 15) - √(4x + 17) = √(x + 2) c, √(x - 2) + √(3 + x) = √(2x + 1) d, √(2x - 1) = √(x - 1) + √(x - 4)

Đáp án:

\(\eqalign{ & a)\,\,x = 2 \cr & b)\,\,x = - 2 \cr & c)\,\,x = 2 \cr & d)\,\,x = 5 \cr} \)

Giải thích các bước giải:

$$\eqalign{ & a)\,\,\sqrt {2x - 1} + \sqrt {x - 2} = \sqrt {x + 1} \,\,\left( {x \ge 2} \right) \cr & \Leftrightarrow 2x - 1 + x - 2 + 2\sqrt {\left( {2x - 1} \right)\left( {x - 2} \right)} = x + 1 \cr & \Leftrightarrow 2x - 4 + 2\sqrt {\left( {2x - 1} \right)\left( {x - 2} \right)} = 0 \cr & \Leftrightarrow \sqrt {\left( {2x - 1} \right)\left( {x - 2} \right)} = 2 - x\,\,\,\left( * \right) \cr & Do\,\,x \ge 2 \Leftrightarrow 2 - x \le 0 \cr & \left( * \right) \Rightarrow 2 - x = 0 \Leftrightarrow x = 2 \cr & Thu\,\,lai: \cr & \sqrt {2.2 - 1} + \sqrt {2 - 2} = \sqrt {2 + 1} \cr & \Leftrightarrow \sqrt 3 = \sqrt 3 \,\,\left( {dung} \right) \cr & Vay\,\,x = 2 \cr & b)\,\,\sqrt {3x + 15} - \sqrt {4x + 17} = \sqrt {x + 2} \,\,\left( {x \ge - 2} \right) \cr & \Leftrightarrow {{3x + 15 - 4x - 17} \over {\sqrt {3x + 15} + \sqrt {4x + 17} }} = \sqrt {x + 2} \cr & \Leftrightarrow {{x + 2} \over {\sqrt {3x + 15} + \sqrt {4x + 17} }} + \sqrt {x + 2} = 0 \cr & \Leftrightarrow \sqrt {x + 2} \left( {{{\sqrt {x + 2} } \over {\sqrt {3x + 15} + \sqrt {4x + 17} }} + 1} \right) = 0 \cr & \Leftrightarrow x + 2 = 0 \Leftrightarrow x = - 2\,\,\left( {tm} \right) \cr & Vay\,\,x = - 2 \cr & c)\,\,\sqrt {x - 2} + \sqrt {3 + x} = \sqrt {2x + 1} \,\,\left( {x \ge 2} \right) \cr & \Leftrightarrow x - 2 + 3 + x + 2\sqrt {\left( {x - 2} \right)\left( {3 + x} \right)} = 2x + 1 \cr & \Leftrightarrow 2\sqrt {\left( {x - 2} \right)\left( {3 + x} \right)} = 0 \cr & \Leftrightarrow \left[ \matrix{ x = 2\,\,\left( {tm} \right) \hfill \cr x = - 3\,\,\left( {ktm} \right) \hfill \cr} \right. \cr & Vay\,\,x = 2 \cr & d)\,\,\sqrt {2x - 1} = \sqrt {x - 1} + \sqrt {x - 4} \,\,\left( {x \ge 4} \right) \cr & \Leftrightarrow 2x - 1 = x - 1 + x - 4 + 2\sqrt {\left( {x - 1} \right)\left( {x - 4} \right)} \cr & \Leftrightarrow \sqrt {\left( {x - 1} \right)\left( {x - 4} \right)} = 2 \cr & \Leftrightarrow \left( {x - 1} \right)\left( {x - 4} \right) = 4 \cr & \Leftrightarrow {x^2} - 5x + 4 = 4 \cr & \Leftrightarrow {x^2} - 5x = 0 \Leftrightarrow \left[ \matrix{ x = 0\,\,\left( {ktm} \right) \hfill \cr x = 5\,\,\left( {tm} \right) \hfill \cr} \right. \cr & Vay\,\,x = 5 \cr} $$