Giải bài 96 trang 132 SGK giải tích nâng cao 12

a) Điều kiện: $x>y>0$

$\left\{ \begin{aligned} & {{\log }_{2}}\left( x-y \right)=5-{{\log }_{2}}\left( x+y \right) \\ & \dfrac{\log x-\log 4}{\log y-\log 3}=-1 \\ \end{aligned} \right. \\ \Leftrightarrow \left\{ \begin{aligned} & {{\log }_{2}}\left( x-y \right)+{{\log }_{2}}\left( x+y \right)=5 \\ & \log x-\log 4+\log y-\log 3=0 \\ \end{aligned} \right. \\ \Leftrightarrow \left\{ \begin{aligned} & {{x}^{2}}-{{y}^{2}}=32 \\ & xy=12 \\ \end{aligned} \right. \\ \Leftrightarrow \left\{ \begin{aligned} & x=\dfrac{12}{y} \\ & \dfrac{144}{{{y}^{2}}}-32-{{y}^{2}}=0 \\ \end{aligned} \right. \\ \Leftrightarrow \left\{ \begin{aligned} & x=\dfrac{12}{y} \\ & {{y}^{4}}+32{{y}^{2}}-144=0 \\ \end{aligned} \right. \\ \Leftrightarrow \left\{ \begin{aligned} & x=\dfrac{12}{y} \\ & y=\pm 2 \\ \end{aligned} \right.$

Do $x>y>0$  $\Rightarrow S=\left\{ \left( 6;2 \right) \right\}$.

b) Điều kiện: $x>0$

$\left\{ \begin{aligned} & 2{{\log }_{2}}x-{{3}^{y}}=15 \\ & {{3}^{y}}.{{\log }_{2}}x=2{{\log }_{2}}x+{{3}^{y+1}} \\ \end{aligned} \right. \\ \Leftrightarrow \left\{ \begin{aligned} & {{3}^{y}}=2{{\log }_{2}}x-15 \\ & \left( 2{{\log }_{2}}x-15 \right){{\log }_{2}}x=2{{\log }_{2}}x+3\left( 2{{\log }_{2}}x-15 \right) \\ \end{aligned} \right. \\ \Leftrightarrow \left\{ \begin{aligned} & {{3}^{y}}=2{{\log }_{2}}x-15 \\ & 2\log _{2}^{2}x-23{{\log }_{2}}x+45=0 \\ \end{aligned} \right. \\ \Leftrightarrow \left\{ \begin{aligned} & {{3}^{y}}=2{{\log }_{2}}x-15 \\ & \left[ \begin{aligned} & {{\log }_{2}}x=9 \\ & {{\log }_{2}}x=2,5 \\ \end{aligned} \right. \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & \left\{ \begin{aligned} & {{3}^{y}}=3 \\ & x={{2}^{9}} \\ \end{aligned} \right. \\ & \left\{ \begin{aligned} & {{3}^{y}}=-10 \\ & x={{2}^{2,5}} \\ \end{aligned} \right.\,\left( \text{loại} \right) \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & y=1 \\ & x=512 \\ \end{aligned} \right.$