\(\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{1}{8}\right)\left(1-\dfrac{1}{9}\right)=\dfrac{a}{75}\\ =>\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{5}{6}\cdot\dfrac{6}{7}\cdot\dfrac{7}{8}\cdot\dfrac{8}{9}=\dfrac{a}{75}\\ =>\dfrac{3\cdot\left(4\cdot5\cdot6\cdot7\cdot8\right)}{\left(4\cdot5\cdot6\cdot7\cdot8\right)\cdot9}=\dfrac{a}{75}\\ =>\dfrac{3}{9}=\dfrac{a}{75}\\ =>\dfrac{a}{75}=\dfrac{1}{3}\\ =>a=\dfrac{1}{3}\cdot75\\ =>a=25\)

Vậy: ...

Số thứ nhất là:

(24680+2):2=24682:2=12341

Số thứ hai là:

12341-2=12339

a)
\(1875:2+125:2\\ =1875\times\dfrac{1}{2}+125\times\dfrac{1}{2}\\ =\dfrac{1}{2}\times\left(1875+125\right)\\ =\dfrac{1}{2}\times2000\\ =1000\)

b)
\(0:36\times\left(32+17+99-68+1\right)\\ =0\times\left(32+17+99-68+1\right)\\ =0\)

c)
\(\left(m:1-m\times1\right):\left(m\times2009+m+1\right)\\ =\left(m-m\right):\left(m\times2009+m+1\right)\\ =0:\left(m\times2009+m+1\right)\\ =0\)

A. 1875 : 2 + 125 : 2

= (1875 + 125) : 2

= 2000 : 2

= 1000

B. 0 : 36 × (32 + 17 + 99 - 68 + 1)

= 0 × (32 + 17 + 99 - 68 + 1)

= 0

C. (m : 1 - m × 1) : (m × 2009 + m + 1)

= (m - m) : (m × 2009 + m + 1)

= 0 : (m × 2009 + m + 1)

= 0

a)
\(8.8.64.2.2\\ =\left(8.2\right).\left(8.2\right).64\\ =16.16.64\\ =4^2.4^2.4^3\\ =4^{2+2+3}\\ =4^7\)

b)
\(27.9.9.3.3\\ =27.\left(9.3\right).\left(9.3\right)\\ =27.27.27\\ =27^{1+1+1}\\ =27^3\)

a) 8.8.64.2.2

= 2³.2³.2⁵.2²

= 2³⁺³⁺⁵⁺²

= 2¹³

b) 27.9.9.3.3

= 3³.3².3².3²

= 3³⁺²⁺²⁺²

= 3⁹

a) 100.10.2.5

= 10².10.10

= 10²⁺¹⁺¹

= 10⁴

b) 16.4.4.2.2.8

= 2⁴.2².2².2².2³

= 2⁴⁺²⁺²⁺²⁺³

= 2¹³

a)
\(100.10.2.5\\ =10^2.10.\left(2.5\right)\\ =10^2.10.10\\ =10^{2+1+1}\\ =10^4\)

b)
\(16.4.4.2.2.8\\ =2^4.2^2.2^2.2.2.2^3\\ =2^{4+2+2+1+1+3}\\ =2^{13}\\ ^{ }\\ \\ \\ \\ \)

1) Ta có:

∠xOn + ∠mOn = 180⁰ (kề bù)

⇒ ∠xOn = 180⁰ - ∠mOn

= 180⁰ - 130⁰

= 50⁰

2) Ta có:

∠xOt + ∠xOn = 180⁰ (kề bù)

⇒ ∠xOt = 180⁰ - ∠xOn

= 180⁰ - 60⁰

= 120⁰

∠tOm = ∠xOn = 60⁰ (đối đỉnh)

∠mOn = ∠xOt = 120⁰ (đối đỉnh)

________, I decided to stop trading with them

A.Despite of the fact that they were the biggest dealer

B.Though being the biggest dealer

C.Being the biggest dealer

D.Even though they were the biggest dealer

Bổ sung: Điều kiện n nguyên

Ta có:
\(12⋮n-1\)
Mà n nguyên nên n-1 nguyên suy ra:
\(n-1\inƯ\left(12\right)\)
Vì \(Ư\left(12\right)=\left\{\pm1,\pm2,\pm3,\pm4,\pm6,\pm12\right\}\) nên:
\(n-1\in\left\{\pm1,\pm2,\pm3,\pm4,\pm6,\pm12\right\}\)
\(\Rightarrow n\in\left\{2;0;3;-1;4;-2;5;-3;7;-5;13;-11\right\}\) (thoả mãn điều kiện)

Vậy \(n\in\left\{2;0;3;-1;4;-2;5;-3;7;-5;13;-11\right\}\)

12 ⋮ (n - 1)

⇒ n - 1 ∈ Ư(12) = {-12; -6; -4; -3; -2; -1; 1; 2; 3; 4; 6; 12}

⇒ n ∈ {-11; -5; -3; -2; -1; 0; 2; 3; 4; 5; 7; 13}