Giải phương trình
$\log _3 \frac{3}{x} .\log_2 x -\log _3 \frac{x^3}{\sqrt{3}} = \frac{1}{2} +\log _2 \sqrt{x}$
Điều kiện $x>0.$
       $\log _3 \frac{3}{x} .\log_2 x -\log _3 \frac{x^3}{\sqrt{3}} = \frac{1}{2} +\log _2 \sqrt{x}$
$\Leftrightarrow \left ( 1-\log_3 x \right ) .\log_2 x -\left (3\log_3 x-\log_3 3^\frac{1}{2} \right )= \frac{1}{2} + \frac{1}{2}\log _2 x$
$\Leftrightarrow \left ( 1-\log_3 x \right ) .\log_2 x -\left (3\log_3 x- \frac{1}{2} \right )= \frac{1}{2} + \frac{1}{2}\log _2 x$
$\Leftrightarrow \log_3 x .\log_2 x - \frac{1}{2}\log _2 x+3\log_3 x=0$
$\Leftrightarrow \frac{\ln x}{\ln 3} . \frac{\ln x}{\ln 2} - \frac{1}{2} \frac{\ln x}{\ln 2}+3 \frac{\ln x}{\ln 3}=0$
$\Leftrightarrow \frac{1}{\ln 3\ln 2}\ln^2 x+\ln x\left ( \frac{3}{\ln 3}-\frac{1}{2\ln 2}\right )=0$
PT này là PT bậc hai theo $\ln x$  có 2 nghiệm ,bạn tự giải tiếp nhé

bai nay rat hay –  nhutuyet12t7.1995 06-12-12 12:58 AM
vote cho bác nhé –  thanhlongntl92 10-10-12 08:39 AM
Điều kiện $x>0$
Đặt $x =3^t$ PT đã cho $\Leftrightarrow \log _3 3^(1-t) . \log _2 3^t  - \log _3 3^(3t-\frac{1}{2}) = \frac{1}{2} + \log _2 3^\frac{t}{2}$
$t(1-t) \log _2 3  -  3t + \frac{1}{2} = \frac{1}{2} +\frac{t}{2} \log _2 3 $
$ \Leftrightarrow t(1-t) \log _2 3  -  3t  = \frac{t}{2} \log _2 3 $
$t((1-t)\log _2 3 -3 - \frac{1}{2} \log _2 3) =0 $
-Nếu $t =0 $ thì $x=1$
-Nghiệm còn lại bạn tựi giai nhé

bai nay thiếu nghiệm ạ –  loannguyen631999 20-11-16 05:57 AM
nhin vao bai thay nen theo huong giai hon la dat an phu –  nhutuyet12t7.1995 06-12-12 01:04 AM
minh thay k on voi loi giai nay .lay co so nao de dat t?trong khi de bai cho nhu the –  nhutuyet12t7.1995 06-12-12 01:03 AM

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