Đáp án:
$\begin{array}{l}
\dfrac{a}{b} > 1 \Leftrightarrow \dfrac{a}{b} > \dfrac{{a + 2001}}{{b + 2001}}\\
\dfrac{a}{b} < 1 \Leftrightarrow \dfrac{a}{b} < \dfrac{{a + 2001}}{{b + 2001}}\\
\dfrac{a}{b} = 1 \Leftrightarrow \dfrac{a}{b} = \dfrac{{a + 2001}}{{b + 2001}}
\end{array}$
Giải thích các bước giải:
ĐKXĐ: $a,b \in Z;b>0$
Ta có:
Xét hiệu:
$\begin{array}{l}
a\left( {b + 2001} \right) - \left( {a + 2001} \right)b\\
= ab + 2001a - ab - 2001b\\
= 2001\left( {a - b} \right)
\end{array}$
+) Nếu $a>b$ khi đó:
$\begin{array}{l}
a\left( {b + 2001} \right) - \left( {a + 2001} \right)b > 0\\
\Leftrightarrow a\left( {b + 2001} \right) > \left( {a + 2001} \right)b\\
\Leftrightarrow \dfrac{a}{b} > \dfrac{{a + 2001}}{{b + 2001}}
\end{array}$
+) Nếu $a<b$ khi đó:
$\begin{array}{l}
a\left( {b + 2001} \right) - \left( {a + 2001} \right)b < 0\\
\Leftrightarrow a\left( {b + 2001} \right) < \left( {a + 2001} \right)b\\
\Leftrightarrow \dfrac{a}{b} < \dfrac{{a + 2001}}{{b + 2001}}
\end{array}$
+) Nếu $a=b$ khi đó:
$\begin{array}{l}
a\left( {b + 2001} \right) - \left( {a + 2001} \right)b = 0\\
\Leftrightarrow a\left( {b + 2001} \right) = \left( {a + 2001} \right)b\\
\Leftrightarrow \dfrac{a}{b} = \dfrac{{a + 2001}}{{b + 2001}}
\end{array}$
Vậy
$\begin{array}{l}
\dfrac{a}{b} > 1 \Leftrightarrow \dfrac{a}{b} > \dfrac{{a + 2001}}{{b + 2001}}\\
\dfrac{a}{b} < 1 \Leftrightarrow \dfrac{a}{b} < \dfrac{{a + 2001}}{{b + 2001}}\\
\dfrac{a}{b} = 1 \Leftrightarrow \dfrac{a}{b} = \dfrac{{a + 2001}}{{b + 2001}}
\end{array}$
