$a$) $A = 3x^2 + 2x - 1$
$TH1$.$x = \dfrac{1}{3}$
Thay $x = \dfrac{1}{3}$ vào $A$, ta được:
$A = 3.(\dfrac{1}{3})^2 + 2.\dfrac{1}{3} - 1$
$A = \dfrac{1}{3}+ \dfrac{2}{3} -1$
$A = 0$
$TH2$.$x= \dfrac{-1}{3}$
Thay $x = \dfrac{-1}{3}$ vào $A$, ta được:
$A = 3.(\dfrac{-1}{3})^2 + 2.\dfrac{-1}{3} - 1$
$A = \dfrac{1}{3}+ \dfrac{-2}{3} -1$
$A = \dfrac{-1}{3} -1$
$A = \dfrac{-4}{3}$
$b$) $B = 3x^2y + 6x^2y^2 +3xy^2$
Thay $x=\dfrac{1}{2};y=\dfrac{-1}{3}$ vào $B$, ta được:
$B = 3.(\dfrac{1}{2})^2 .\dfrac{-1}{3} + 6.(\dfrac{1}{2})^2.(\dfrac{-1}{3})^2 + 3.\dfrac{1}{2}.(\dfrac{-1}{3})^2$
$B = 3.\dfrac{1}{4} . \dfrac{-1}{3} + 6.\dfrac{1}{4} . \dfrac{1}{9} + 3.\dfrac{1}{2}.\dfrac{1}{9}$
$B = \dfrac{-1}{4} + \dfrac{1}{6} + \dfrac{1}{6}$
$B = \dfrac{1}{12}$
$B = -1 + \dfrac{2}{3} + \dfrac{1}{6}$
$B = -1 + \dfrac{5}{6}$
$B = \dfrac{-1}{6}$
