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

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

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

a) Ta có:

`m^2>=0` với mọi m 

`=>m^2+1/2>=1/2>0` với mọi m 

`=>` Bất pt: `(m^2+1/2)x-1<=0` có hệ số `a≠0` 

`=>`Bất pt luôn là bất pt bậc nhất 1 ẩn với mọi m 

b) Ta có:

`m^2+m+2=(m^2+2*m*1/2+1/4)+7/4` 

`=(m+1/2)^2+7/4>=7/4>=0` với mọi m

`=>-(m^2+m+2)<=-7/2<0` với mọi m

`=>-(m^2+m+2)≠0` với mọi m 

=> Bất pt `-(m^2+m+2)x<=-m+2024` luôn là bpt bậc nhất 1 ẩn 

\(4x^2-25+\left(2x+5\right)^2=0\\ < =>\left[\left(2x\right)^2-5^2\right]+\left(2x+5\right)^2=0\\ < =>\left(2x+5\right)\left(2x-5\right)+\left(2x+5\right)^2=0\\ < =>\left(2x+5\right)\left(2x-5+2x+5\right)=0\\ < =>4x\left(2x+5\right)=0\\ < =>\left[{}\begin{matrix}4x=0\\2x+5=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\2x=-5\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)

Vậy: ... 

`24^2 - 25 + (2x + 5)^2 = 0`

Ta có: `24^2 > 25`

`=> 24^2 - 25 > 0`

Và `(2x + 5)^2 >= 0 ∀x `

`=> 24^2 - 25 + (2x + 5)^2 > 0`

Vậy phương trình đã cho vô nghiệm

`-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 ...

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

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

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

\(=43.200\)

\(=8600\)

Đề là gì bạn?

`124 . (-49) + 62 . (-102)`

`= 124 . (-49) + 62 . 2.(-51) `

`=124. (-49) +124. (-102)`

`= 124 . (-49 + (-102))`

`= 124 . (-100)`

`= -12400`

\(=124.\left(-49\right)+62.2.\left(-51\right)\)

\(=124.\left(-49\right)+124.\left(-51\right)\)

\(=124.\left(-49+\left(-51\right)\right)\)

\(=124.\left(-100\right)\)

\(=-12400\)