Ta có: \(|a|\) = \(\dfrac{1}{3}\) \(\Rightarrow\) a = \(\pm\)\(\dfrac{1}{3}\)

\(|b|\) = 0,25 \(\Rightarrow\) b= \(\pm\) 0,25

Ta xét hai trường hợp:

* Với a = \(\dfrac{1}{3}\) và b = 0,25 thì:

a) A = 3a - 3ab - b

= 3.\(\dfrac{1}{3}\) - 3.\(\dfrac{1}{3}\). 0,25 - 0,25

= 1-\(\dfrac{1}{4}\)- 0,25

= \(\dfrac{1}{2}\)

b) B = 5.\(\dfrac{a}{3}\) -3b

= 5.\(\dfrac{1}{\dfrac{3}{3}}\) - 3.0,25

= \(\dfrac{5}{9}\) -\(\dfrac{3}{4}\)

= \(\dfrac{-7}{36}\)

* Với a = -\(\dfrac{1}{3}\) , b = -0,25 thì:

a) A = 3a - 3ab -b

= 3.(-\(\dfrac{1}{3}\))- 3.(-\(\dfrac{1}{3}\) ).(-0,25) -(-0,25)

= -1 - 0,25 - 0,25

= -\(\dfrac{3}{2}\)

b) B = 5.\(\dfrac{a}{3}\) - 3b

= 5.\(\dfrac{-1}{\dfrac{3}{3}}\) - 3. (-0,25)

= \(\dfrac{-5}{9}\) - (-\(\dfrac{3}{4}\) )

= \(\dfrac{7}{36}\)

1)

a) x > 0 \(\Leftrightarrow\) a-3 \(\in\) \(\left\{2;4;6;8;...\right\}\)

\(\Leftrightarrow\)a \(\in\) \(\left\{5;7;9;11;...\right\}\)

Vậy a\(\in\) \(\left\{5;7;9;11;...\right\}\)

b) x < 0 \(\Leftrightarrow\) a-3 \(\in\left\{-4;-6;-8;-10;...\right\}\)

\(\Leftrightarrow\) a\(\in\left\{-1;-3;-5;-7;-9;...\right\}\)

Vậy a\(\in\left\{-1;-3;-5;-7;-9;-11;...\right\}\)

c) x = 0 \(\Leftrightarrow\) ( a-3 ) : 2 = 0

\(\Rightarrow\) a-3 = 0.2

\(\Rightarrow\) a-3 = 0

\(\Rightarrow\) a = 0+3

\(\Rightarrow\) a = 3

Vậy a=3

Mk k chắc lắm!!

Ủa thế rốt cuộc là tính j z bn?