Đáp án:
$$m \in \left[ {1;10} \right]$$
Giải thích các bước giải:
$$\eqalign{
& \sqrt {x + 3 + 4\sqrt {x - 1} } + \sqrt {x + 8 - 6\sqrt {x - 1} } = 5\,\,\left( {x \ge 1} \right) \cr
& \Leftrightarrow \sqrt {x - 1 + 4\sqrt {x - 1} + 4} + \sqrt {x - 1 - 6\sqrt {x - 1} + 9} = 5 \cr
& \Leftrightarrow \sqrt {{{\left( {\sqrt {x - 1} + 2} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 3} \right)}^2}} = 5 \cr
& \Leftrightarrow \sqrt {x - 1} + 2 + \left| {\sqrt {x - 1} - 3} \right| = 5 \cr
& TH1:\,\,\sqrt {x - 1} - 3 \ge 0 \Leftrightarrow \sqrt {x - 1} \ge 3 \Leftrightarrow x - 1 \ge 9 \Leftrightarrow x \ge 10 \cr
& \Leftrightarrow \sqrt {x - 1} + 2 + \sqrt {x - 1} - 3 = 5 \cr
& \Leftrightarrow 2\sqrt {x - 1} = 6 \cr
& \Leftrightarrow \sqrt {x - 1} = 3 \Leftrightarrow x - 1 = 9 \Leftrightarrow x = 10\,\,\left( {tm} \right) \cr
& TH2:\,\,\sqrt {x - 1} - 3 < 0 \Leftrightarrow 1 \le x < 10 \cr
& \Leftrightarrow \sqrt {x - 1} + 2 - \sqrt {x - 1} + 3 = 5 \cr
& \Leftrightarrow 5 = 5\,\,\left( {luon\,\,dung} \right) \cr
& Vay\,\,m \in \left[ {1;10} \right] \cr} $$
