Số vịt có trong trang trại là:

`200:2=100` (con) 

Tổng số ngan và ngỗng là:

`200:2=100` (con) 

Số ngan có trong trang trại là: 

`(100+40):2=70` (con) 

Số ngỗng có trong vườn là:

`100-70=30` (con)

ĐS: ... 

\(a,k-13=7\\ k=13+7\\ k=20\\ b,7\times x+5=19\\ 7\times x=19-5\\ 7\times x=14\\ x=14:7\\ x=2\\ c,28:\left(3\times c-2\right)=7\\ 3\times c-2=28:7\\ 3\times c-2=4\\ 3\times c=2+4\\ 3\times c=6\\ c=6:3\\ c=2\\ d,x+0,25=4:\dfrac{3}{8}\\ x+0,25=4\times\dfrac{8}{3}\\ x+\dfrac{1}{4}=\dfrac{32}{3}\\ x=\dfrac{32}{3}-\dfrac{1}{4}\\ x=\dfrac{125}{12}\) 

\(e,96-3,75:x=94,5\\ 3,75:x=96-94,5\\ 3,75:x=1,5\\ x=3,75:1,5\\ x=2,5\\ f,\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{5}{6}\\ \dfrac{1}{4}:x=\dfrac{5}{6}-\dfrac{3}{4}\\ \dfrac{1}{4}:x=\dfrac{1}{12}\\ x=\dfrac{1}{4}:\dfrac{1}{12}=3\\ g,\dfrac{6}{13}:\left(\dfrac{1}{2}+x\right)=\dfrac{15}{39}\\ \dfrac{1}{2}+x=\dfrac{6}{13}:\dfrac{15}{39}\\ \dfrac{1}{2}+x=\dfrac{6}{5}\\ x=\dfrac{6}{5}-\dfrac{1}{2}\\ x=\dfrac{7}{10}\\ h,x\times1,75+x\times3,5+x\times4,75=15\\ x\times\left(1,75+3,5+4,75\right)=15\\ x\times10=15\\ x=\dfrac{15}{10}\\ x=\dfrac{3}{2}\)

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

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}