Lời giải

a) \(\left\{ \begin{array}{l}\frac{1}{{x - 2}} + \frac{1}{{y - 1}} = 2\\\frac{2}{{x - 2}} - \frac{3}{{y - 1}} = 1\end{array} \right.\) (I)

Điều kiện: \(\left\{ \begin{array}{l}x \ne 2\\y \ne 1\end{array} \right.\) (*)

Đặt \(a = \frac{1}{{x - 2}};\,b = \frac{1}{{y - 1}}\).

Hệ (I) tương đương với: \(\left\{ \begin{array}{l}a + b = 2\\2a - 3b = 1\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}a = 2 - b\\2\left( {2 - b} \right) - 3b = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 2 - b\\ - 5b = - 3\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}a = 2 - b\\b = \frac{3}{5}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \frac{7}{5}\\b = \frac{3}{5}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}\frac{1}{{x - 2}} = \frac{7}{5}\\\frac{1}{{y - 1}} = \frac{3}{5}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x - 2 = \frac{5}{7}\\y - 1 = \frac{5}{3}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}x = \frac{{19}}{7}\\y = \frac{8}{3}\end{array} \right.\)

So với điều kiện (*), ta nhận \(\left\{ \begin{array}{l}x = \frac{{19}}{7}\\y = \frac{8}{3}\end{array} \right.\)

Vậy \(\left( {x;y} \right) \in \left\{ {\left( {\frac{{19}}{7};\frac{8}{3}} \right)} \right\}\).

b) \(\left\{ \begin{array}{l}\frac{3}{{2x - 1}} - \frac{6}{{3 - y}} = - 1\\\frac{1}{{2x - 1}} - \frac{3}{{3 - y}} = 0\end{array} \right.\) (II)

Điều kiện: \(\left\{ \begin{array}{l}x \ne \frac{1}{2}\\y \ne 3\end{array} \right.\) (*)

Đặt \(a = \frac{1}{{2x - 1}};\,b = \frac{1}{{3 - y}}\).

Hệ (II) tương đương với: \(\left\{ \begin{array}{l}3a - 6b = - 1\\a - 3b = 0\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}3a - 6b = - 1\\a = 3b\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}9b - 6b = - 1\\a = 3b\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}3b = - 1\\a = 3b\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = - \frac{1}{3}\\a = - 1\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}\frac{1}{{3 - y}} = - \frac{1}{3}\\\frac{1}{{2x - 1}} = - 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}3 - y = - 3\\2x - 1 = - 1\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}y = 6\\x = 0\end{array} \right.\)

So với điều kiện (*), ta nhận \(\left\{ \begin{array}{l}y = 6\\x = 0\end{array} \right.\)

Vậy (x; y) ∈ {(0; 6)}.

c) \(\left\{ \begin{array}{l}\frac{1}{x} + \frac{1}{y} = \frac{1}{4}\\\frac{{10}}{x} + \frac{1}{y} = 1\end{array} \right.\) (III)

Điều kiện: x, y ≠ 0 (*)

Đặt \(a = \frac{1}{x};\,\,b = \frac{1}{y}\).

Hệ (III) tương đương với: \(\left\{ \begin{array}{l}a + b = \frac{1}{4}\\10a + b = 1\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}a + b = \frac{1}{4}\\b = 1 - 10a\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a + 1 - 10a = \frac{1}{4}\\b = 1 - 10a\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l} - 9a = - \frac{3}{4}\\b = 1 - 10a\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \frac{1}{{12}}\\b = \frac{1}{6}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}\frac{1}{x} = \frac{1}{{12}}\\\frac{1}{y} = \frac{1}{6}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 12\\y = 6\end{array} \right.\)

So với điều kiện (*), ta nhận \(\left\{ \begin{array}{l}x = 12\\y = 6\end{array} \right.\)

Vậy (x; y) ∈ {(12; 6)}.

d) \(\left\{ \begin{array}{l}\frac{2}{{2x - y}} + \frac{3}{{x - 2y}} = \frac{1}{2}\\\frac{2}{{2x - y}} - \frac{1}{{x - 2y}} = \frac{1}{{18}}\end{array} \right.\) (IV)

Điều kiện: \(\left\{ \begin{array}{l}x \ne 2y\\x \ne \frac{y}{2}\end{array} \right.\) (*)

Đặt \(a = \frac{1}{{2x - y}};\,b = \frac{1}{{x - 2y}}\).

Hệ (IV) tương đương với: \(\left\{ \begin{array}{l}2a + 3b = \frac{1}{2}\\2a - b = \frac{1}{{18}}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}2a + 3b = \frac{1}{2}\\b = 2a - \frac{1}{{18}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2a + 3\left( {2a - \frac{1}{{18}}} \right) = \frac{1}{2}\\b = 2a - \frac{1}{{18}}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}8a = \frac{2}{3}\\b = 2a - \frac{1}{{18}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \frac{1}{{12}}\\b = \frac{1}{9}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}\frac{1}{{2x - y}} = \frac{1}{{12}}\\\frac{1}{{x - 2y}} = \frac{1}{9}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2x - y = 12\\x - 2y = 9\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}y = 2x - 12\\x - 2\left( {2x - 12} \right) = 9\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = 2x - 12\\ - 3x = - 15\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}y = - 2\\x = 5\end{array} \right.\)

So với điều kiện (*), ta nhận \(\left\{ \begin{array}{l}y = - 2\\x = 5\end{array} \right.\)

Vậy (x; y) ∈ {(5; –2)}.

Phân tích các đa thức sau thành nhân tử (x^2 + 1)^2 – 4x^2