Giải PT: $2^{\sqrt{x^2+1}}\log_2 (x+\sqrt{x^2+1})=4^xlog_2(3x)$
với x>0 phương trình tương đương

$2^{x+\sqrt{x^{2}+1}}log_{2}(x+\sqrt{x^{2}+1})=2^{3x}log_{2}3x$

đặt $f(t)=2^{t}log_{2}t$ (t>0)

$f'(t)=2^{t}(ln2.log_{2}t+\frac{1}{tln2})$

nếu t$\geq 1$ thì $f'(t)>0$

nếu 0<t<1 ta xét $g(t)=\frac{f'(t)}{2^{t}}=ln2.log_{2}t+\frac{1}{tln2}$

$g'(t)=\frac{1}{t}-\frac{1}{t^{2}ln2}=\frac{t^{2}ln2-t}{t^{3}ln2}<0$

nên $g(t) $ nghịch biến suy ra $g(t)>g(1)=\frac{1}{ln2}>0\Rightarrow f'(t)>0$

từ đó suy ra $f(t)$ đồng biến $\forall t>0$

mà $f(x+\sqrt{x^{2}+1})=f(3x)\Rightarrow \sqrt{x^{2}+1}=2x$

dễ dàng giải pt này ta được nghiệm $x=\frac{1}{\sqrt{3}}$
@@! chăm gõ quá! –  Confusion 23-04-16 07:49 PM

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