2.$log_2(x)=log_3(\sqrt{x}+2)$ đặt $t=log_2(x) ->x=2^t$ thay vào ta được $t=log_3(\sqrt{2^t}+2)$ $<-> 3^t=\sqrt{2^t}+2$ $<-> 3^t-\sqrt{2^t}=2$ xét hàm số $f(t)=3^t-\sqrt{2^t}$ 1. $<->3^{x^2}.2^{\frac{x}{2x-1}}=6$ $<->log_63^{x^2}+log_62^{\frac{x}{2x-1}}=1$ $<->x^2.log_63+\frac{x}{2x-1}log_62=1$ $<->log_63.2x^3-log_63.x^2+log_62.x-1=0$ $<->x=1$