Bất phương trình (sqrt {x – 1} > sqrt {x – 2} + sqrt {x – 3} ) có bao nhiêu nghiệm nguyên dương?

ĐKXĐ: (left{ begin{array}{l}x – 1 ge 0\x – 2 ge 0\x – 3 ge 0end{array} right.)( Leftrightarrow left{ begin{array}{l}x ge 1\x ge 2\x ge 3end{array} right.)( Leftrightarrow x ge 3)

(begin{array}{l}sqrt {x – 1}  > sqrt {x – 2}  + sqrt {x – 3} \ Leftrightarrow x – 1 > x – 2 + x – 3 + 2sqrt {left( {x – 2} right)left( {x – 3} right)} \ Leftrightarrow 2sqrt {left( {x – 2} right)left( {x – 3} right)}  < 4 – x\ Leftrightarrow left{ begin{array}{l}4 – x > 0\left( {x – 2} right)left( {x – 3} right) ge 0\4left( {x – 2} right)left( {x – 3} right) < {left( {4 – x} right)^2}end{array} right. Leftrightarrow left{ begin{array}{l}x < 4\left[ begin{array}{l}x le 2\x ge 3end{array} right.\4{x^2} – 20x + 24 < 16 – 8x + {x^2}end{array} right.\ Leftrightarrow left{ begin{array}{l}x < 4\left[ begin{array}{l}x le 2\x ge 3end{array} right.\3{x^2} – 12x + 8 < 0end{array} right. Leftrightarrow left{ begin{array}{l}left[ begin{array}{l}x le 2\3 le x < 4end{array} right.\dfrac{{6 – 2sqrt 3 }}{3} < x < dfrac{{6 + 2sqrt 3 }}{3}end{array} right.\ Leftrightarrow x in left( {dfrac{{6 – 2sqrt 3 }}{3};2} right] cup left[ {3;dfrac{{6 – 2sqrt 3 }}{3}} right)end{array})

Mà (x in {mathbb{Z}^ + }) và (x ge 3) nên (x = 3).

Chọn B.