Bài 3:

Để A là số nguyên thì \(2n+7⋮n+3\)

=>\(2n+6+1⋮n+3\)

=>\(1⋮n+3\)

=>\(n+3\in\left\{1;-1\right\}\)

=>\(n\in\left\{-2;-4\right\}\)

Bài 4:

Để A là số nguyên thì \(2n+7⋮n+1\)

=>\(2n+2+5⋮n+1\)

=>\(5⋮n+1\)

=>\(n+1\in\left\{1;-1;5;-5\right\}\)

=>\(n\in\left\{0;-2;4;-6\right\}\)

Bài 5:

Để A là số nguyên thì \(6n-3⋮3n+1\)

=>\(6n+2-5⋮3n+1\)

=>\(-5⋮3n+1\)

=>\(3n+1\in\left\{1;-1;5;-5\right\}\)

=>\(3n\in\left\{0;-2;4;-6\right\}\)

=>\(n\in\left\{0;-\dfrac{2}{3};\dfrac{4}{3};-2\right\}\)

mà n nguyên 

nên \(n\in\left\{0;-2\right\}\)

Bài 6:
Để A là số nguyên thì \(3n+4⋮n-1\)

=>\(3n-3+7⋮n-1\)

=>\(7⋮n-1\)

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

=>\(n\in\left\{2;0;8;-6\right\}\)

Câu 1:

\(M=132-\left\{100-\left[\left(78-73\right)^2:5+9\right]\right\}\)

\(=132-\left\{100-\left[5^2:5+9\right]\right\}\)

\(=132-100+5+9\)

=32+14

=46

Câu 2:

105-[(2x+7)-13]=25

=>(2x+7)-13=105-25=90

=>2x-6=90

=>2x=96

=>x=96/2=48

Câu 4:

\(N=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}\)

Câu 5:
\(xy-x+2y=3\)
\(\Rightarrow xy-x+2y-2=3-2\)
\(\Rightarrow\left(xy-x\right)+\left(2y-2\right)=1\)
\(\Rightarrow x\left(y-1\right)+2\left(y-1\right)=1\)
\(\Rightarrow\left(x+2\right)\left(y-1\right)=1\)
Ta có bảng sau:

Vậy các cặp (x;y) là (-1;2) ; (-3;0)
#YM

B

B

b

Bài 4:

a) 2x + 7 ⋮ x + 2

⇒ 2x + 4 + 3 ⋮ x + 2

⇒ 2(x + 2) + 3 ⋮ x + 2

⇒ 3 ⋮ x + 2

⇒ x + 2 ∈ Ư(3) = {1; -1; 3; -3}

⇒ x ∈ {-1; -3; 1; -5} 

b) 2x + 7 ⋮ x - 3

⇒ 2x - 6 + 13 ⋮ x - 3

⇒ 2(x - 3) + 13 ⋮ x - 3

⇒ 13 ⋮ x - 3

⇒ x - 3 ∈ Ư(13) = {1; -1; 13; -13}

⇒ x ∈ {4; 2; 16; -10}

Bài 6:

a: \(3x-13⋮x+3\)

=>\(3x+9-22⋮x+3\)

=>\(-22⋮x+3\)

=>\(x+3\in\left\{1;-1;2;-2;11;-11;22;-22\right\}\)

=>\(x\in\left\{-2;-4;-1;-5;8;-14;19;-25\right\}\)

b: \(2x+24⋮x-4\)

=>\(2x-8+32⋮x-4\)

=>\(32⋮x-4\)

=>\(x-4\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16;32;-32\right\}\)

=>\(x\in\left\{5;3;6;2;8;0;12;-4;20;-12;36;-28\right\}\)

Bài 5:

a: \(4x+3⋮x-2\)

=>\(4x-8+11⋮x-2\)

=>\(11⋮x-2\)

=>\(x-2\in\left\{1;-1;11;-11\right\}\)

=>\(x\in\left\{3;1;13;-9\right\}\)

b: \(2x+7⋮x-3\)

=>\(2x-6+13⋮x-3\)

=>\(13⋮x-3\)

=>\(x-3\in\left\{1;-1;13;-13\right\}\)

=>\(x\in\left\{4;2;16;-10\right\}\)

Bài 4:

a: \(2x+7⋮x+2\)

=>\(2x+4+3⋮x+2\)

=>\(3⋮x+2\)

=>\(x+2\in\left\{1;-1;3;-3\right\}\)

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

b: \(2x+7⋮x-3\)

=>\(2x-6+13⋮x-3\)

=>\(13⋮x-3\)

=>\(x-3\in\left\{1;-1;13;-13\right\}\)

=>\(x\in\left\{4;2;16;-10\right\}\)

a) 2x + 7 ⋮ x + 2

⇒ 2x + 4 + 3 ⋮ x + 2

⇒ 2(x + 2) + 3 ⋮ x + 2

⇒ 3 ⋮ x + 2

⇒ x + 2 ∈ Ư(3) = {1; -1; 3; -3}

⇒ x ∈ {-1; -3; 1; -5} 

b) 2x + 7 ⋮ x - 3

⇒ 2x - 6 + 13 ⋮ x - 3

⇒ 2(x - 3) + 13 ⋮ x - 3

⇒ 13 ⋮ x - 3

⇒ x - 3 ∈ Ư(13) = {1; -1; 13; -13}

⇒ x ∈ {4; 2; 16; -10}

38A  39B  40C

28D

29C

30C

31B

Xét ΔABC có BC^2=AB^2+AC^2

nên ΔABC vuông tại A

Xét (B;BA) có

BA là bán kính

CA vuông góc BA tại A

Do đó: CA là tiếp tuyến của (B;BA)