$\eqalign{
& \sqrt {x + 2} + \sqrt {6 - x} + \sqrt {\left( {x + 2} \right)\left( {6 - x} \right)} = m \cr
& dk: - 2 \le x \le 6 \cr
& dieu\;kien\;can:\;xet\;f\left( x \right) = \sqrt {x + 2} + \sqrt {6 - x} + \sqrt {\left( {x + 2} \right)\left( {6 - x} \right)} \cr
& gia\;su\;tim\;duoc\;m\;thoa\;man\;de\;bai\;khi\;do\;phuong\;trinh\;co\;nghiem\;duy\;nhat\;la\;{x_0}\;\left( { - 2 \le {x_0} \le 6} \right) \cr
& \to \sqrt {{x_0} + 2} + \sqrt {6 - {x_0}} + \sqrt {\left( {{x_0} + 2} \right)\left( {6 - {x_0}} \right)} = m \cr
& ta\;co\;f\left( {4 - {x_0}} \right) = \sqrt {6 - {x_0}} + \sqrt {2 + {x_0}} + \sqrt {\left( {6 - {x_0}} \right)\left( {{x_0} + 2} \right)} = m \cr
& va\; - 2 \le 4 - {x_0} \le 6 \cr
& \to 4 - {x_0}\;cung\;la\;mot\;nghiem\;cua\;pt \cr
& \to phuong\;trinh\;co\;nghiem\;duy\;nhat \leftrightarrow {x_0} = 4 - {x_0} \leftrightarrow {x_0} = 2 \to m = 8 \cr
& dieu\;kien\;du:\;voi\;m = 8\;thay\;vao\;giai\;xem\;co\;nghiem\;duy\;nhat\;khong\;roi\;ket\;luan \cr} $
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