Phương trình logx 2 + log2 x = 5/2 có 2 nghiệm x1, x2 (x1 < x2)

ĐKХĐ: \(x > 0\,;\,\,x \ne 1\).

Ta có: \({\log _x}2 + {\log _2}x = \frac{5}{2} \Leftrightarrow \frac{1}{{{{\log }_2}x}} + {\log _2}x = \frac{5}{2}.\)

Đặt \(t = {\log _2}x \Rightarrow \frac{1}{t} + t = \frac{5}{2} \Rightarrow 1 + {t^2} - \frac{5}{2}t = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{t = 2}\\{t = \frac{1}{2}}\end{array}} \right..\)

Suy ra: \[\left[ {\begin{array}{*{20}{l}}{{{\log }_2}x = 2}\\{{{\log }_2}x = \frac{1}{2}}\end{array} \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{{x_2} = 4}\\{{x_1} = \sqrt 2 }\end{array}} \right.} \right.\] (vì \(\left. {{x_1} < {x_2}} \right)\) \( \Rightarrow x_1^2 + {x_2} = 2 + 4 = 6.\) Chọn C.