tìm số nguyên x;y (x-3)(x-5)<0 xy-2x+3y=1 2^x+3+2^x+4=192 2^x.3^x=4.3^2

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\left( {x – 3} \right)\left( {x – 5} \right) < 0\\
TH1:\,\,x \le 3 \Rightarrow \left\{ \begin{array}{l}
x – 3 \le 0\\
x – 5 < 0
\end{array} \right. \Rightarrow \left( {x – 3} \right)\left( {x – 5} \right) \ge 0\\
TH2:\,\,\,x \ge 5 \Rightarrow \left\{ \begin{array}{l}
x – 3 > 0\\
x – 5 \ge 0
\end{array} \right. \Rightarrow \left( {x – 3} \right)\left( {x – 5} \right) \ge 0\\
TH3:\,\,3 < x < 5 \Rightarrow \left\{ \begin{array}{l}
x – 3 > 0\\
x – 5 < 0
\end{array} \right. \Rightarrow \left( {x – 3} \right)\left( {x – 5} \right) < 0\\
x \in Z \Rightarrow x = 4\\
b,\\
xy – 2x + 3y = 1\\
 \Leftrightarrow x\left( {y – 2} \right) + 3\left( {y – 2} \right) = 1 – 6\\
 \Leftrightarrow \left( {x + 3} \right)\left( {y – 2} \right) =  – 5\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 3 = 1\\
y – 2 =  – 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 3 =  – 1\\
y – 1 = 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 3 =  – 5\\
y – 1 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
y + 3 = 5\\
y – 1 =  – 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x =  – 2\\
y =  – 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x =  – 4\\
y = 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x =  – 8\\
y = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 2\\
y = 0
\end{array} \right.
\end{array} \right.\\
c,\\
{2^{x + 3}} + {2^{x + 4}} = 192\\
 \Leftrightarrow {2^{x + 3}}\left( {1 + 2} \right) = 192\\
 \Leftrightarrow {3.2^{x + 3}} = 192\\
 \Leftrightarrow {2^{x + 3}} = 64\\
 \Leftrightarrow x + 3 = 6\\
 \Leftrightarrow x = 3\\
d,\\
{2^x}{.3^x} = {4.3^2}\\
 \Leftrightarrow {6^x} = {6^2}\\
 \Leftrightarrow x = 2
\end{array}\)