Bài 1: Tính:

a, (-8) × 25 × (-2) × 4 × (-5) × 125

= (-8 x 125) x (25 x 4) x [-2 x (-5)]

= -1000 x 100 x 10

= -1000000

b,19 × 25 + 9 × 95 + 19 × 30

= 19 x (25 + 30) + 9 x 95

= 19 x 55 + 9 x 95

= 1045 + 855

= 1900

c, (-48) + 48 × (-78) + 48 × (-21)

= 48 x (-1 - 78 - 21)

= 48 x (-100)

= -4800

d, 13 × (25 + 22) - 3 x (17 + 18)

= 13 x 47 - 3 x 35 

= 611 - 105

= 506

e, 29 × (19 - 13) - 19 × (29 - 13)

= 29 x 6 - 19 x 16

= 174 - 304

= -130

f, 17 × (-57) - 23 × 37 - 46 × (-37)

= -969 - 851 + 1702

= -118

g, (-37 - 17) × (-9) + 37 × (-9 - 11)

= -54 x (-9) + 37 x (-20)

= 486 - 740

= -254

h, (-25) × (75 + 45) - 75 × (45 - 25)

= -25 x 120 - 75 x 20

= -3000 - 1500

= -4500

Bài 2: Tìm x thuộc Z:

a, (x + 3) × (x - 4) = 0

\(\left[ \begin{array}{l}x+3=0\\x-4=0\end{array} \right.\)

\(\left[ \begin{array}{l}x=0\\x=4\end{array} \right.\)

Vậy x = 0 hoặc x = 4

b, (x - 1) × (x - 3) = 0

\(\left[ \begin{array}{l}x-1=0\\x-3=0\end{array} \right.\)

\(\left[ \begin{array}{l}x=1\\x=3\end{array} \right.\)

Vậy x = 1 hoặc x = 3

Đáp án:

Giải thích các bước giải:

Bài 1: Tính:

a,(-8) × 25 × (-2) × 4 × (-5) × 125

=[(-2) × (-5)]× (25 × 4) × [(-8) × 125]

=10 × 100 × (-1000)

= (-1000000)

b,19 × 25 + 9 × 95 + 19 × 30

=19 × 25 + 9 × 5 × 19+ 19 × 30

=19 × 25 + 45 × 19+ 19 × 30

=19 × (25 + 45 + 30)

=19 × 100

=1900

c,(-48) + 48 × (-78) + 48 × (-21)

=(-1) × 48+ 48 × (-78) + 48 × (-21)

=(-1 - 78 - 21) × 48

=(-100) × 48

=(-4800)

d,13 × (25 + 22) - 3 × (17 + 18)

=13 × 47 - 3 × 35

=611 - 105

=506

e,29 × (19 - 13) - 19 × (29 - 13)

=29 × 19 - 29 × 13 - 19 × 29 + 19 × 13

=(29 × 19 - 19 × 29) + (19 × 13 - 29 × 13)

=0 + 13 × (19 - 29)

=13 × (-10)

=(-130)

f,17 × (-57) - 23 × 37 - 46 × (-37)

=(-969) - 851 - (-1702)

=(-969) - 851 + 1702

=(-118)

g,(-37 - 17) × (-9) + 37 × (-9 - 11)

=(-54)× (-9) + 37 × (-20) 

=486 + (-740)

=486 - 740

=(-254)

h,(-25) × (75 + 45) - 75 × (45 - 25)

=(-25) × 120 - 75 × 20

=(-3000) - 1500

=(-4500)

Bài 2:Tìm x thuộc Z:

a,(x + 3) × (x - 4)=0

⇔\(\left[ \begin{array}{l}x + 3=0\\x - 4=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x =-3\\x =4\end{array} \right.\) 

Vậy x ∈ {-3;4} (tmđk x ∈ Z)

b,(x - 1) × (x - 3)=0

⇔\(\left[ \begin{array}{l}x - 1=0\\x - 3=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x =1\\x =3\end{array} \right.\) 

Vậy x ∈ {1;3} (tmđk x ∈ Z)