Bài 2:
a) (x - 29) - 11 = 0
⇔ x - 29 = 0 + 11
⇔ x - 29 = 11
⇔ x = 11 + 29
⇔ x = 40
Vậy x = 40
b) 531 - (312 - x) + 15 = 246
⇔ 531 - (312 - x) = 246 - 15
⇔ 531 - (312 - x) = 231
⇔ 312 - x = 531 - 231
⇔ 312 - x = 300
⇔ x = 312 - 300
⇔ x = 12
Vậy x = 12
c) (4x + 5) : 3 - 121 : 11 = 4
⇔ (4x + 5) : 3 - 11 = 4
⇔ (4x + 5) : 3 = 4 + 11
⇔ (4x + 5) : 3 = 15
⇔ 4x + 5 = 15 . 3
⇔ 4x + 5 = 45
⇔ 4x =45 - 5
⇔ 4x = 40
⇔ x = 40 : 4
⇔ x = 10
Vậy x = 10
d) 131x - 941 = $2^{7}$ . $2^{3}$
⇔ 131x - 941 = $2^{7 + 3}$
⇔ 131x - 941 = $2^{10}$
⇔ 131x - 941 = 1024
⇔ 131x = 1024 + 941
⇔ 131x = 1965
⇔ x = 1965 : 131
⇔ x = 15
Vậy x = 15
e) 225 : 15 + 3 . (2x + 1) = 270
⇔ 15 + 3 . (2x + 1) = 270
⇔ 3 . (2x + 1) = 270 - 15
⇔ 3 . (2x + 1) = 255
⇔ 2x + 1 = 255 : 3
⇔ 2x + 1 = 83
⇔ 2x = 83 - 1
⇔ 2x = 82
⇔ x = 82 : 2
⇔ x = 41
Vậy x = 41
f) 19 . $(2 + 3 + 4 - 5 + 6 - 7)^{2}$ - 9 . (7x - 2) = 0
⇔ 19 . $3^{2}$ - 9 . (7x - 2) = 0
⇔ 19 . 9 - 9 . (7x - 2) = 0
⇔ 171 - 9 . (7x - 2) = 0
⇔ 9 . (7x - 2) = 171 - 0
⇔ 9 . (7x - 2) = 171
⇔ 7x - 2 = 171 : 9
⇔ 7x - 2 = 19
⇔ 7x = 19 + 2
⇔ 7x = 21
⇔ x = 21 : 3
⇔ x = 7
Vậy x = 7
g) 3 . $(2x + 1)^{3}$ = 81
⇔ $(2x + 1)^{3}$ = 81 : 3
⇔ $(2x + 1)^{3}$ = 27
⇔ $(2x + 1)^{3}$ = $3^{3}$
⇒ 2x + 1 = 3
⇔ 2x = 3 - 1
⇔ 2x = 2
⇔ x = 2 : 2
⇔ x = 1
Vậy x = 1
h) $(x + 1)^{5}$ = 243
⇔ $(x + 1)^{5}$ = $3^{5}$
⇒ x + 1 = 3
⇔ x = 3 - 1
⇔ x = 2
Vậy x = 2
i) 2 . $11^{x}$ = $(3^2 + 2)^{3}$ : ($5^{3}$ - $2^{5}$ : $2^{3}$) . 22
⇔ 2 . $11^{x}$ = $(9 + 2)^{3}$ : (125 - 4) . 22
⇔ 2 . $11^{x}$ = $11^{3}$ : 121 . 22
⇔ 2 . $11^{x}$ = $11^{3}$ : $11^{2}$ . 22
⇔ 2 . $11^{x}$ = 11 . 22
⇔ 2 . $11^{x}$ = 242
⇔ $11^{x}$ = 242 : 2
⇔ $11^{x}$ = 121
⇔ $11^{x}$ = $11^{2}$
⇒ x = 2
Vậy x = 2
k) $7^{x}$ + $7^{x+1}$ + $7^{x+2}$ = 3 . 19 . 343
⇔ $7^{x}$ + $7^{x+1}$ + $7^{x+2}$ = 19551
⇔ $7^{x}$ + $7^{x+1}$ . $7^{1}$ + $7^{x+2}$ . $7^{2}$ = 19551
⇔ $7^{x}$ . (1 + $7^{1}$ + $7^{2}$) = 19551
⇔ $7^{x}$ . (1 + 7 + 49) = 19551
⇔ $7^{x}$ . 57 = 19551
⇔ $7^{x}$ = 19551 : 57
⇔ $7^{x}$ = 343
⇔ $7^{x}$ = $7^{3}$
⇒ x = 3
Vậy x = 3
Bài 4:
a) (1 . 3 . 5 . 7 .9) `\vdots` 5; 20 `\vdots` 5
⇒ A `\vdots` 5; A > 5. Vậy A là hợp số.
b) (21 . 22 . 23) `\vdots` 7; (6 . 7 . 8) `\vdots` 7
⇒ B `\vdots` 7; A > 7. Vậy B là hợp số.