- Cách 1
$4-(\frac{3}{2})^x-18.(\frac{9}{4})^x=0$
Đặt $$(\frac{3}{2})^x=a$$ Ta có :
$4-a-18.a^2=0\Leftrightarrow 18a^2+a-4=0\Leftrightarrow \begin{bmatrix} a_1=\frac{4}{9} & & \\ a_2=-0.5(KTM) & & \end{bmatrix}$
Vậy $a=\frac{4}{9}\Leftrightarrow (\frac{3}{2})^x=\frac{4}{9}=(\frac{2}{3})^2=(\frac{3}{2})^{-2}$
Vậy $ x=-2$
- Cách 2
$4.(2^{x})^{2}-2^{x}.3^{x}-18.(3^{x})^{2}=0$
Đặt $2^{x}=a, 3^{x}=b$
$\Rightarrow 4a^{2}-ab+18b^{2}=0$
$\Rightarrow a=\frac{9b}{4}$ ( nhận) hay $a=-2b$ (loại)
Nếu $a=\frac{9}{4}b$
$\Rightarrow 2^{x}=\frac{9}{4}.3^{x}$
$\Rightarrow (\frac{2}{3})^{x}=\frac{9}{4}$
$\Rightarrow x=-2$
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