a, $ 3^{x+1}+ 18. 3^{-x}=29 $
b,$ 27^x +12^x = 2. 8^x $
Chắc mất căn bản
Câu a.
Pt $\Leftrightarrow 3.3^x+\frac{18}{3^x}=29$
$\Leftrightarrow 3.(3^x)^2-29.3^x+18=0$
Đặt $t=3^x>0$
Pt trở thành: $3t^2-29t+18=0$
$\Leftrightarrow t=9\vee t=\frac{2}{3}$
Với $t=3^x=9=3^2\Leftrightarrow x=2$
Với $t=3^x=\frac{2}{3}\Leftrightarrow x=\log_3\frac{2}{3}$
Câu b. Chia cho cơ số nhỏ nhất hoặc lớn nhất
Ta chia cả 2 vế cho $8^x$ thì được
$(\frac{3^3}{2^3})^x+(\frac{12}{8})^x-2=0$
$\Leftrightarrow [(\frac{3}{2})^x]^3+(\frac{3}{2})^x-2=0$
$\Leftrightarrow (\frac{3}{2})^x=1\Leftrightarrow x=0$

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