a, Ta có: x ∈ B(5) = {0;5;10;15;20;25;...}

Mà x < 20 => x ∈ {0;5;10;15}

b, Ta có: x ⋮ 24; x ⋮ 45 => x ∈ BC{24;45}

24 = 2 3 . 3 ; 45 = 3 2 . 5 => BCNN(24;45) =  2 3 . 3 2 . 5 = 360

=> BC(24;45) = B(360) = {0;360;720;...}

Mà 200 < x < 500 => x = 360

c, Ta có:

452 = px+32 => 420 ⋮ x

321 = qx+21 => 320 ⋮ x

=> x ∈ ƯC(420;320)

420 =  2 2 . 3 . 5 . 7

320 =  2 6 . 5

=> ƯCLN(420;320) = 20 =  2 2 . 5

=> ƯC(420;320) = Ư(20) = {1;2;4;5;10;20}

Vậy x ∈ {1;2;4;5;10;20}

\(=321-21\cdot\left[2\cdot27+256:32-52\right]\)

\(=321-21\cdot10=321-210=111\)

321 x 3 + 321 x 5 + 321 x 2

= 321 x (3 + 5 + 2)

= 321 x 10

= 3 210.

321 x 3 + 321 x 5 + 321 x 2

= 321 x (3 + 5 + 2)

= 321 x 10

= 3210

a)   \(\left|x\right|+2=2^3.\left(-2\right)^2\)

\(\Leftrightarrow\)\(\left|x\right|+2=32\)

\(\Leftrightarrow\)\(\left|x\right|=30\)

\(\Leftrightarrow\)\(x=\pm30\)

Vậy...

b) \(\left|x\right|+4=\left(-3\right)^2+\left(-2\right)\)

\(\Leftrightarrow\)\(\left|x\right|+4=7\)

\(\Leftrightarrow\)\(\left|x\right|=3\)

\(\Leftrightarrow\)\(x=\pm10\)

Vậy...

b2 :

a) 5 . ( -5 - 6 )

= 5 . ( -11 )

= -55

b) 321 + { - 15 + [ 30 + ( -321 ) ] }

= 321 + { - 15 + [ - 291 ] }

= 321 + ( - 306 )

= 15

19 . 92 - 9 . ( - 19 ) - 19

= 1748 - ( -171 ) - 19

= 1919 - 19

= 1900

321- 21. [(2.3^3+4^4 :32)]- 52

=321- 21. [(2.27+256 :32)]- 52

=321-21.[(54+8)]-52

=321-21.62-52

=321-1203-52

=-754

giusp mik câu dưới với

b: =>x-4=-4

hay x=0

a/|x|-2,5=27,5

=>|x|=27,5+2,5=30

=>x=30 hoặc x=-30

b/\(\dfrac{3}{4}+\dfrac{2}{5}.x=\dfrac{29}{60}\)

=>\(\dfrac{2}{5}.x\)=\(\dfrac{29}{60}-\dfrac{3}{4}\)=\(\dfrac{-4}{15}\)

=>x=\(\dfrac{-4}{15}:\dfrac{2}{5}\)=\(\dfrac{-2}{3}\)

c/(x-1)\(^5\)=-32

=>x-1=-2 vì (-2)\(^5\)=-32

=>x=-2+1=-1

d/\(\dfrac{4}{5}.x+0,5=4.5\)

=>\(\dfrac{4}{5}.x+0,5=20\)

=>\(\dfrac{4}{5}.x=20-0,5=19,5\)

=>\(x=19,5:\dfrac{4}{5}\)=\(\dfrac{195}{8}\)