Bài 5 :
$A=4.|x-2|+1 ≥1$
Dấu "=" xảy ra $⇔x-2=0⇔x=2$
Vậy GTNN của $A = 1$ tại $x=2$
$B = |x-2020|+|x-1| $
$=|x-2020|+|1-x| ≥ |x-2020+1-x| = 2019$
Dấu "=" xảy ra $⇔(x-2020).(1-x) ≥0$
$⇔1≤x≤2020$
Bài 2 :
$\dfrac{2}{3}+\dfrac{5}{6}.x=\dfrac{1}{2}$
$⇔ \dfrac{5}{6}.x = \dfrac{-1}{6}$
$⇔x=\dfrac{-1}{6}:\dfrac{5}{6} = \dfrac{-1}{5}$
Bài 1 :
a) $3\dfrac{1}{2} + \dfrac{-4}{7} :\dfrac{8}{9} $
$=\dfrac{7}{2} + (\dfrac{-9}{14} ) $
$=\dfrac{20}{7}$
b) $\dfrac{3}{4}.\dfrac{-5}{11}+\dfrac{5}{11}.\dfrac{-1}{4}$
$=\dfrac{5}{11}.(\dfrac{-3}{4}+\dfra{-1}{4}) $
$=\dfrac{5}{11}.(-1) = \dfrac{-5}{11}$
