Bài 96 trang 132 SGK giải tích 12 nâng cao

\(\left\{ \matrix{
{\log _2}\left( {x - y} \right) = 5 - {\log _2}\left( {x + y} \right) \hfill \cr 
{{\log x - \log 4} \over {\log y - \log 3}} = - 1 \hfill \cr} \right.\)

Lời giải chi tiết:

Điều kiện: 

\(\left\{ \matrix{
x > 0;\,y > 0 \hfill \cr 
x - y > 0;\,x + y > 0 \hfill \cr} \right. \Leftrightarrow x > y > 0\)

Khi đó,

\(\left\{ \matrix{
{\log _2}\left( {x - y} \right) = 5 - {\log _2}\left( {x + y} \right) \hfill \cr 
{{\log x - \log 4} \over {\log y - \log 3}} = - 1 \hfill \cr} \right.\)

\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
{\log _2}\left( {x - y} \right) + {\log _2}\left( {x + y} \right) = 5\\
\log x - \log 4 = - \log y + \log 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\log _2}\left[ {\left( {x - y} \right)\left( {x + y} \right)} \right] = 5\\
\log x + \log y = \log 3 + \log 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\log _2}\left( {{x^2} - {y^2}} \right) = 5\\
\log \left( {xy} \right) = \log 12
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} - {y^2} = {2^5} = 32\\
xy = 12
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = \frac{{12}}{x}\\
{x^2} - \frac{{144}}{{{x^2}}} = 32
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = \frac{{12}}{x}\\
{x^4} - 32x - 144 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = \frac{{12}}{x}\\
\left[ \begin{array}{l}
{x^2} = 36\\
{x^2} = - 4\left( {loai} \right)
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = \frac{{12}}{x}\\
\left[ \begin{array}{l}
x = 6\left( {TM} \right)\\
x = - 6\left( {loai} \right)
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 6\\
y = 2
\end{array} \right.\left( {TM} \right)
\end{array}\)

Vậy \(S = \left\{ {\left( {6;2} \right)} \right\}\)

\(\left\{ \matrix{
2{\log _2}x - {3^y} = 15 \hfill \cr 
{3^y}.{\log _2}x = 2{\log _2}x + {3^{y + 1}} \hfill \cr} \right.\)

Lời giải chi tiết:

Điều kiện: \(x > 0\).

Đặt \(\left\{ \begin{array}{l}
u = {\log _2}x\\
v = {3^y}>0
\end{array} \right.\) ta có hệ phương trình:

\(\left\{ \matrix{
2u - v = 15\,\,\,\,\left( 1 \right) \hfill \cr 
u.v = 2u + 3v\,\,\,\,\left( 2 \right) \hfill \cr} \right.\)

Từ (1) suy ra \(v = 2u – 15\), thay vào (2) ta được:

\(\eqalign{
& u\left( {2u - 15} \right) = 2u + 3\left( {2u - 15} \right) \cr&\Leftrightarrow 2{u^2} - 23u + 45 = 0 \cr 
& \Leftrightarrow \left\{ \matrix{
u = 9 \hfill \cr 
u = {5 \over 2} \hfill \cr} \right. \cr} \)

Với \(u = 9 \Rightarrow v = 2.9 - 15 = 3\left( {TM} \right)\)

Với \(u = \frac{5}{2} \Rightarrow v = 2.\frac{5}{2} - 15 =  - 10\left( {loai} \right)\)

Vậy 

\(\left\{ \matrix{
u = 9 \hfill \cr 
v = 3 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
\log _2x = 9 \hfill \cr 
{3^y} = 3 \hfill \cr} \right. \)

\(\Leftrightarrow \left\{ \matrix{
x = {2^9} = 512 \hfill \cr 
y = 1 \hfill \cr} \right.\)

Vậy \(S = \left\{ {\left( {512;1} \right)} \right\}\)