Bài 1:
a) Ta có: $\dfrac{17}{30} = \dfrac{17.3}{30.3} = \dfrac{51}{90}$
Do $90 < 92$
nên $\dfrac{1}{90} > \dfrac{1}{92}$
$\Rightarrow \dfrac{51}{90} > \dfrac{51}{92}$
Vậy $\dfrac{17}{30} > \dfrac{51}{92}$
b) Ta có: $\dfrac{-3}{5} = \dfrac{-3.3}{5.3} = \dfrac{-9}{15}$
Do $15 < 23$
nên $\dfrac{1}{15} > \dfrac{1}{23}$
$\Rightarrow \dfrac{-3}{15} < \dfrac{-3}{23}$
Vậy $\dfrac{-3}{5} < \dfrac{-9}{23}$
Bài 2:
a) Ta có: $\dfrac{1313}{2727} = \dfrac{13.101}{27.101} = \dfrac{13}{27}$
Vậy $\dfrac{13}{27} = \dfrac{1313}{2727}$
b) Ta có:$\dfrac{-151515}{232323} = \dfrac{-15.10101}{23.10101} = \dfrac{-15}{23}$
Vậy $\dfrac{-15}{23} = \dfrac{-151515}{232323}$
Bài 3:
a) Ta có: $|x + \dfrac{3}{2}| \geq 0, \forall x$
Dấu "=" xảy ra $\Leftrightarrow x + \dfrac{3}{2} = 0 \Leftrightarrow x = -\dfrac{3}{2}$
Vậy $minA = 0$ tại $x = -\dfrac{3}{2}$
b) Ta có:$ |x + \dfrac{1}{2}| \geq 0, \forall x$
$\Leftrightarrow |x + \dfrac{1}{2}| + \dfrac{3}{4} \geq \dfrac{3}{4}, \forall x$
Dấu "=" xảy ra $\Leftrightarrow x + \dfrac{1}{2} = 0 \Leftrightarrow x = -\dfrac{1}{2}$
Vậy $minB = \dfrac{3}{4}$ tại $x = -\dfrac{1}{2}$
