√3x 1 - √6-x 3x^2 - 14x - 8 = 0
Giải thích
Lời giải: ĐKXĐ: \(\frac{{ - 1}}{3} \le x \le 6\).
\(\sqrt {3{\rm{x}} + 1} - \sqrt {6 - x} + 3{{\rm{x}}^2} - 14{\rm{x}} - 8 = 0\)
\(\left( {\sqrt {3{\rm{x}} + 1} - 4} \right) - \left( {\sqrt {6 - x} - 1} \right) + 3{{\rm{x}}^2} - 14{\rm{x}} - 5 = 0\)
\(\frac{{3{\rm{x}} - 15}}{{\sqrt {3{\rm{x}} + 1} + 4}} + \frac{{x - 5}}{{\sqrt {6 - x} + 1}} + \left( {x - 5} \right)\left( {3{\rm{x}} + 1} \right) = 0\)
\(\left( {x - 5} \right)\left( {\frac{3}{{\sqrt {3{\rm{x}} + 1} }} + \frac{1}{{\sqrt {6 - x} + 1}} + 3{\rm{x}} + 1} \right) = 0\)
Do \(\frac{{ - 1}}{3} \le x \le 6\)nên 3x + 1 ≥0
Suy ra \(\frac{3}{{\sqrt {3{\rm{x}} + 1} }} + \frac{1}{{\sqrt {6 - x} + 1}} + 3{\rm{x}} + 1\)>0
Suy ra x = 5.