Đáp án:
a, $\dfrac{-2003}{2004}>$\dfrac{-2005}{2003}$
b, $\dfrac{11}{20}>\dfrac{23}{47}$
Giải thích các bước giải:
a, $\dfrac{-2005}{2003}= \dfrac{-2003-2}{2003}=\dfrac{-2003}{2003}-\dfrac{2}{2003}$
mà $\dfrac{-2003}{2003}<\dfrac{-2003}{2004}$
⇒ $\dfrac{-2003}{2004}>\dfrac{-2003}{2003}-\dfrac{2}{2003}$ =$\dfrac{-2005}{2003}$
b, $\dfrac{11}{20}=\dfrac{517}{940}$ ; $\dfrac{23}{47}=\dfrac{460}{940}$
$\dfrac{517}{940}>\dfrac{460}{940}$ ⇒ $\dfrac{11}{20}>\dfrac{23}{47}$
