\(\dfrac{4}{9\cdot11}+\dfrac{4}{13\cdot15}+...+\dfrac{4}{95\cdot97}+\dfrac{4}{97\cdot99}\\ =2\cdot\left(\dfrac{2}{9\cdot11}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{95\cdot97}+\dfrac{2}{97\cdot99}\right)\\ =2\left(\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}\right)\\ =2\cdot\left(\dfrac{1}{9}-\dfrac{1}{99}\right)\\ =2\cdot\dfrac{11-1}{99}\\ =2\cdot\dfrac{10}{99}\\ =\dfrac{20}{99}\)

Sửa đề: `S = 4/(9.11) + 4/(11.13) + ... + 4/(97.99)`

`S = 2 . (2/(9.11) + 2/(11.13) + ... +2/(97.99))`

`S = 2 . (1/9 - 1/11 + 1/11 - 1/13 + ... + 1/97 - 1/99)`

`S = 2 . (1/9 - 1/99)`

`S = 2 . (11/99 - 1/99)`

`S = 2 . 10/99 `

`S = 20/99`

`527 + {[2 . (2 . 2^3 + 3^2 + 4^2 - 5^2) + 678^0]^3 : 33^2}`

`= 527 + {[2 . (16 + 9 + 16 - 25) + 1]^3 : 33^2}`

`= 527 + {[2 . (25 + 16 - 25) + 1]^3 : 33^2}`

`= 527 + {[2 . 16  + 1]^3 : 33^2}`

`= 527 + {[32  + 1]^3 : 33^2}`

`= 527 + {33^3 :33^2}`

`= 527 + 33^(3-2)`

`= 527 + 33`

`= 560`

`(42 . 43 + 43 . 57 + 43) - 360 : 4 `

`= 43 . (42 + 57 + 1) - 90`

`= 43 . 100 - 90`

`= 4300 - 90 `

`= 4210`

`372 - 19 . 4 + (981:9-13)`

`= 372 - 76 + (109-13)`

`= 296 + 96`

`= 392`

M là tập hợp các số tự nhiên lẻ không lớn hơn 11 

Cách 1: liệt kê

\(M=\left\{1;3;5;7;9;11\right\}\)

Cách 2: chỉ ra tính chất đặt trưng

`M={x=2k+1;k∈N;0<=k<=5}` 

Ta có: 13 không có trong tập hợp M 

`=>13∉M`

9 có trong tập hợp M 

`=>9∈M`

C2 sai rồi

a) `(x+4)(y-1)=13` 

Ta có bảng: 

b) `xy-3x+y=20`

`=>(xy-3x)+(y-3)=20-3`

`=>x(y-3)+(y-3)=17`

`=>(y-3)(x+1)=17` 

Ta có bảng:

a) 10 ⋮ (x - 1)

⇒ x - 1 ∈ Ư(10) = {-10; -5; -2; -1; 1; 2; 5; 10}

⇒ x ∈ {-9; -4; -1; 0; 2; 3; 6; 11}

b) x + 5 = x - 2 + 7

Để (x - 5) ⋮ (x - 2) thì 7 ⋮ (x - 2)

⇒ x - 2 ∈ Ư(7) = {-7; -1; 1; 7}

⇒ x ∈ {-5; 1; 3; 9}

c) 3x + 8 = 3x - 3 + 11

= 3(x - 1) + 11

Để (3x + 8) ⋮ (x - 3) thì 11 ⋮ (x - 3)

⇒ x - 3 ∈ Ư(11) = {-11; -1; 1; 11}

⇒ x ∈ {-8; 2; 4; 14}

a) 25 chia hết cho n + 2

=> n + 2 ∈ Ư(25) 

=> n + 2 ∈ {1; -1; 5; -5; 25; -25}

=> n ∈ {-1; -3; 3; -7; 23; -27}  

b) 2n + 4 chia hết cho n - 1

=> (2n - 2) + 6 chia hết chi n - 1

=> 2(n - 1) + 6 chia hết cho n - 1

=> 6 chia hết cho n - 1

=> n - 1 ∈ Ư(6) = {1; -1; 2; -2; 3; -3; 6; -6}

=> n ∈ {2; 0; 3; -1; 4; -2; 7; -5} 

a) 25 ⋮ (n + 2)

⇒ n + 2 ∈ Ư(25) = {-25; -5; -1; 1; 5; 25}

⇒ n ∈ {-27; -7; -3; -1; 3; 23}

b) 2n + 4 = 2n - 2 + 6

= 2(n - 1) + 6

Để (2n + 4) ⋮ (n - 1) thì 6 ⋮ (n - 1)

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

⇒ n ∈ {-5; -2; -1; 0; 2; 3; 4; 7}

Bài 11.2 

\(a,A=3+3^2+3^3+....+3^{99}\\ =\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+....+\left(3^{97}+3^{98}+3^{99}\right)\\ =3\cdot\left(1+3+9\right)+3^4\cdot\left(1+3+9\right)+...+3^{97}\cdot\left(1+3+9\right)\\ =13\cdot\left(3+3^4+....+3^{97}\right)⋮13\\ b,B=5+5^2+...+5^{50}\\ =\left(5+5^2\right)+\left(5^3+5^4\right)+..+\left(5^{49}+5^{50}\right)\\ =5\cdot\left(1+5\right)+5^3\cdot\left(1+5\right)+....+5^{49}\cdot\left(1+5\right)\\ =6\cdot\left(5+5^3+...+5^{49}\right)⋮6\)

Bài 11.4:

a: \(10⋮x-1\)

=>\(x-1\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)

=>\(x\in\left\{2;0;3;-1;6;-4;11;-9\right\}\)

b: 

\(x+5⋮x-2\)

=>\(x-2+7⋮x-2\)

=>\(7⋮x-2\)

=>\(x-2\in\left\{1;-1;7;-7\right\}\)

=>\(x\in\left\{3;1;9;-5\right\}\)

c: \(3x+8⋮x-1\)

=>\(3x-3+11⋮x-1\)

=>\(11⋮x-1\)

=>\(x-1\in\left\{1;-1;11;-11\right\}\)

=>\(x\in\left\{2;0;12;-10\right\}\)

Bài 11.5:

a: (x+4)(y-1)=13

=>\(\left(x+4;y-1\right)\in\left\{\left(1;13\right);\left(13;1\right);\left(-1;-13\right);\left(-13;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(-3;14\right);\left(9;2\right);\left(-5;-12\right);\left(-17;0\right)\right\}\)

b: xy-3x+y=20

=>x(y-3)+y-3=17

=>(x+1)(y-3)=17

=>\(\left(x+1;y-3\right)\in\left\{\left(1;17\right);\left(17;1\right);\left(-1;-17\right);\left(-17;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(0;20\right);\left(16;4\right);\left(-2;-14\right);\left(-18;2\right)\right\}\)

`-12(x - 15) + 7(3 - x) = 15`

`=> -12x + 180 + 21 - 7x - 15 = 0`

`=> -19x + 186 = 0`

`=> -19x = -186`

`=> x = -186 : (-19) `

`=> x = 186/19`

Vậy ...

-------------------------

`(x - 1) . (x + 2) . (-x - 3) = 0`

`=> x - 1 = 0` hoặc `x + 2 = 0` hoặc `-x - 3 = 0`

`=> x =1` hoặc `x = -2` hoặc `x = -3`

Vậy ...

\(=-125.75-125.\left(-43\right)+75.125+75.43\)

\(=\left(-125.75+75.125\right)+125.43+75.43\)

\(=0+43.\left(125+75\right)\)

\(=43.200\)

\(=8600\)