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

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)

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}

Phân số cần tìm là: