Giải bài 72-73 trang 127 SGK giải tích nâng cao 12

$72.a)\,\left\{ \begin{aligned} & x+y=20 \\ & {{\log }_{4}}x+{{\log }_{4}}y=1+{{\log }_{4}}9 \\ \end{aligned} \right.$

$\Leftrightarrow \left\{ \begin{aligned} & x+y=20 \\ & {{\log }_{4}}xy={{\log }_{4}}4+{{\log }_{4}}9 \\ \end{aligned} \right. \\ \Leftrightarrow \left\{ \begin{aligned} & x+y=20 \\ & xy=36 \\ \end{aligned} \right.$

Suy ra x, y là nghiệm của phương trình:

${{X}^{2}}-20X+36=0 \\ \Leftrightarrow \left[ \begin{aligned} & X=18 \\ & X=2 \\ \end{aligned} \right.$

 $\Rightarrow (x;y)\in\left\{ \left( 2;18 \right),\left( 18;2 \right) \right\}$.

$b)\,\left\{ \begin{aligned} & x+y=1 \\ & {{4}^{-2x}}+{{4}^{-2y}}=0,5 \\ \end{aligned} \right. \\ \Leftrightarrow \left\{ \begin{aligned} & y=1-x \\ & {{4}^{-2x}}+{{4}^{-2\left( 1-x \right)}}=\dfrac{1}{2}\,\left( * \right) \\ \end{aligned} \right.$

$\left( * \right)\Leftrightarrow {{4}^{-2x}}+{{4}^{-2+2x}}=\dfrac{1}{2}$

Đặt ${{4}^{2x}}=t$ phương trình trở thành

$\dfrac{1}{t}+\dfrac{t}{16}=\dfrac{1}{2}\Leftrightarrow {{t}^{2}}-8t+16=0 \\ \Leftrightarrow {{\left( t-4 \right)}^{2}}=0\Leftrightarrow t=4 \\ \Leftrightarrow {{4}^{2x}}=4\Leftrightarrow 2x=1 \\ \Leftrightarrow x=\dfrac{1}{2}$

$\Rightarrow y=\dfrac{1}{2}$

Vậy $(x;y)=(1;2)$

73.a) Điều kiện $x + y > 0$

$\left\{ \begin{aligned} & {{3}^{-x}}{{.2}^{y}}=1152 \\ & {{\log }_{\sqrt{5}}}\left( x+y \right)=2 \\ \end{aligned} \right. \Leftrightarrow \left\{ \begin{aligned} & {{3}^{-x}}{{.2}^{y}}=1152 \\ & x+y=5 \\ \end{aligned} \right. \\ \Leftrightarrow \left\{ \begin{aligned} & x=5-y \\ & {{3}^{y-5}}{{.2}^{y}}=1152 \\ \end{aligned} \right. \Leftrightarrow \left\{ \begin{aligned} & x=5-y \\ & {{ 6 }^{y}}={{1152.3}^{5}}={{6}^{7}} \\ \end{aligned} \right.$

$\Leftrightarrow \left\{ \begin{aligned} & y=7 \\ & x=-2 \\ \end{aligned} \right.$ (thỏa mãn)

b) Điều kiện: $\left\{ \begin{aligned} & x+y>0 \\ & x-y>0 \\ \end{aligned} \right.$

$\left\{ \begin{aligned} & {{x}^{2}}-{{y}^{2}}=2 \\ & {{\log }_{2}}\left( x+y \right)-{{\log }_{3}}\left( x-y \right)=1 \\ \end{aligned} \right. \\ \Leftrightarrow \left\{ \begin{aligned} & {{\log }_{2}}\left( x+y \right)+{{\log }_{2}}\left( x-y \right)=1 \\ & {{\log }_{2}}\left( x+y \right)-\dfrac{{{\log }_{2}}\left( x-y \right)}{{{\log }_{2}}3}=1 \\ \end{aligned} \right.$

Đặt $\left\{ \begin{aligned} & {{\log }_{2}}\left( x+y \right)=u \\ & {{\log }_{2}}\left( x-y \right)=v \\ \end{aligned} \right.$ hệ phương trình trở thành

$\left\{ \begin{aligned} & u+v=1 \\ & u-v.{{\log }_{3}}2=1 \\ \end{aligned} \right. \Leftrightarrow \left\{ \begin{aligned} & u=1 \\ & v=0 \\ \end{aligned} \right. \\ \Leftrightarrow \left\{ \begin{aligned} & x+y=2 \\ & x-y=1 \\ \end{aligned} \right. \Leftrightarrow \left\{ \begin{aligned} & x=\dfrac{3}{2} \\ & y=\dfrac{1}{2} \\ \end{aligned} \right.$