Bải 2 $2.1$ Ta có: $(2-3i).\overline{z}-1-i+4i^{2016}=0$$\Leftrightarrow (2-3i).\overline{z}=i-3$$\Leftrightarrow \overline{z}=\frac{i-3}{2-3i}=-\frac{9}{13}-\frac{7i}{13}$$\Rightarrow z=-\frac{9}{13}+\frac{7i}{13}$$\color{red}{\Rightarrow \left| {z} \right|=\sqrt{(-\frac{9}{13})^2+(\frac{7}{13})^2}=\sqrt{\frac{10}{13}}.}$ $2.2$ Ta có: $3.16^x+2.81^x=5.36^x$$\Leftrightarrow 3.(\frac{4}{9})^x+2.(\frac{9}{4})^x=5$ $(\bigstar)$ Đặt $t=(\frac{4}{9})^x,(t>0)$, PT $(\bigstar)$ trở thành: $3t+\frac{2}{t}=5\Leftrightarrow 3t^2-5t+2=0\Leftrightarrow \left[\ \begin{array}{l} t=1\\ t=\frac{2}{3} \end{array} \right.\color{red}{\Leftrightarrow \left[\ \begin{array}{l} x=0\\ x=\frac{1}{2} \end{array} \right.}$
B
ài 2 $2.1$ Ta có: $(2-3i).\overline{z}-1-i+4i^{2016}=0$$\Leftrightarrow (2-3i).\overline{z}=i-3$$\Leftrightarrow \overline{z}=\frac{i-3}{2-3i}=-\frac{9}{13}-\frac{7i}{13}$$\Rightarrow z=-\frac{9}{13}+\frac{7i}{13}$$\color{red}{\Rightarrow \left| {z} \right|=\sqrt{(-\frac{9}{13})^2+(\frac{7}{13})^2}=\sqrt{\frac{10}{13}}.}$ $2.2$ Ta có: $3.16^x+2.81^x=5.36^x$$\Leftrightarrow 3.(\frac{4}{9})^x+2.(\frac{9}{4})^x=5$ $(\bigstar)$ Đặt $t=(\frac{4}{9})^x,(t>0)$, PT $(\bigstar)$ trở thành: $3t+\frac{2}{t}=5\Leftrightarrow 3t^2-5t+2=0\Leftrightarrow \left[\ \begin{array}{l} t=1\\ t=\frac{2}{3} \end{array} \right.\color{red}{\Leftrightarrow \left[\ \begin{array}{l} x=0\\ x=\frac{1}{2} \end{array} \right.}$