Đáp án:
a) $\left( {a;b} \right) \in \left\{ {\left( {6;0} \right);\left( {2;1} \right)} \right\}$
b) $\left( {a;b} \right) \in \left\{ {\left( {7;1} \right);\left( {5;2} \right);\left( {4;4} \right)} \right\}$
Giải thích các bước giải:
a) ĐK: $a\ne 0$
Ta có:
$\begin{array}{l}
\dfrac{1}{a} = \dfrac{1}{6} + \dfrac{b}{3}\\
\Leftrightarrow \dfrac{1}{a} = \dfrac{{1 + 2b}}{6}\\
\Leftrightarrow a\left( {2b + 1} \right) = 6\left( 1 \right)\\
a,b \in N \Rightarrow 2b + 1\not \vdots 2\\
\left( 1 \right) \Rightarrow \left[ \begin{array}{l}
a = 6;2b + 1 = 1\\
a = 2;2b + 1 = 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
a = 6;b = 0\\
a = 2;b = 1
\end{array} \right.
\end{array}$
Vậy $\left( {a;b} \right) \in \left\{ {\left( {6;0} \right);\left( {2;1} \right)} \right\}$
b) ĐK: $b\ne 0$
Ta có:
$\begin{array}{l}
\dfrac{a}{4} - \dfrac{1}{b} = \dfrac{3}{4}\\
\Leftrightarrow \dfrac{{a - 3}}{4} = \dfrac{1}{b}\\
\Leftrightarrow \left( {a - 3} \right)b = 4\left( 2 \right)\\
a,b \in N\\
\left( 2 \right) \Rightarrow \left[ \begin{array}{l}
b = 1;a - 3 = 4\\
b = 2;a - 3 = 2\\
b = 4;a - 3 = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
b = 1;a = 7\\
b = 2;a = 5\\
b = 4;a = 4
\end{array} \right.
\end{array}$
Vậy $\left( {a;b} \right) \in \left\{ {\left( {7;1} \right);\left( {5;2} \right);\left( {4;4} \right)} \right\}$
