[Toán 12] Phương trình mũ

T

truongduong9083

Giải:
Cách 1:
Vì; $\sqrt{x^2+1}>\sqrt{x^2}=|x|\ge x\Longrightarrow x+\sqrt{x^2+1}>0$
Nên:
$$(1)\Longleftrightarrow \ln (x+\sqrt{x^2+1})=x\ln 3$$$$\Longleftrightarrow f(x)=\ln (x+\sqrt{x^2+1})-x\ln 3=0$$
Ta có: $f'(x)=\frac{1}{\sqrt{x^2+1}}-\ln 3<0$ \forall $x\in \mathbb{R}$
$\Longrightarrow x=0$

Cách 2:
Vì $-x+\sqrt{x^2+1} \ne 0$
Nên: $(x+\sqrt{x^2+1})(-x+\sqrt{x^2+1})=3^x(-x+\sqrt{x^2+1}) \Longrightarrow -x+\sqrt{x^2+1}=3^{-x}$
Suy ra hệ: $$ \left\{\begin{array}{1} x+\sqrt{x^2+1}=3^x \\ -x+\sqrt{x^2+1}=3^{-x} \end{array}\right.$$$$
\Longleftrightarrow \left\{\begin{array}{1} 2\sqrt{x^2+1}=3^x+3^{-x} \\
2x=3^x-3^{-x} \end{array}\right.$$$$ \Longrightarrow 4(1+(3^x-3^{-x})^2)=(3^x+3^{-x})^2 $$$$\Longleftrightarrow 4+4(t^2+\frac{1}{t^2}-2)=(t^2+\frac{1}{t^2}+2) \,\ (t=3^x) $$$$\Longleftrightarrow (t^2-1)^2=0$$
 
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