chịu

b, $\frac{n-1}{n-2}$

Ta có: $\frac{\mathrm{\backslash (\backslash dfrac\{n-1\}\{n-2\}\backslash )=\; \backslash (\backslash dfrac\{n-2+3\}\{n-2\}=\backslash dfrac\{n-2\}\{n-2\}+\backslash dfrac\{3\}\{n-2\}=1+\backslash dfrac\{3\}\{n-2\}\backslash )}}{}$

$\frac{\mathrm{Để\; (n-1)\; chia\; hết\; (n-2)\; thì\; 3\; chia\; hết\; cho\; (n-2)}}{}$

$\frac{\mathrm{Hay\; (n-2)\; thuộc\; Ư(3)}}{}$

$\frac{\mathrm{Ta\; có\; :\; Ư(3)=\backslash (\backslash left\backslash \{-3;-1;1;3\backslash right\backslash \}\backslash )}}{}$

$\frac{\mathrm{TH1:\; n-2\; =\; -3\; \backslash (\backslash Rightarrow\; n=-1\backslash )}}{}$

TH2: n-2= -1 \(\Rightarrow n=1\)

TH3: n-2 = 1\(\Rightarrow n=3\)

TH4: n- 2 = 3\(\Rightarrow n=5\)

Vậy n\(\in\left\{-1;1;3;5\right\}\)thì \(\dfrac{n-1}{n-2}\)

Bài 1: 

a: Để A là phân số thì n+1<>0

hay n<>-1

b: Để A là số nguyên thì \(n+1\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{0;-2;4;-6\right\}\)

Bài 1: 

a: Để A là số nguyên thì \(x+1⋮3\)

=>x=3k-1, với k là số nguyên

b; Để B là số nguyên thì \(x-1\in\left\{1;-1;17;-17\right\}\)

hay \(x\in\left\{2;0;18;-16\right\}\)

Bài 1:

a: Để hàm số đồng biến thì a>0

Để hàm số nghịch biến thì a<0

b: Để hai đường vuôg góc thì a*1=-1

=>a=-1

Bài 2:

PTHĐGĐ là:

1/4x^2=2x+m-4

=>x^2=8x+4m-16

=>x^2-8x-4m+16=0

Δ=(-8)^2-4(-4m+16)

=64+16m-64=16m

Để (P) cắt (d) tại hai điểm phân biệt thì 16m>0

=>m>0

a) Để A có giá trị nguyên

\(\Rightarrow\frac{3}{x-1}\in z\Rightarrow3⋮x-1\)

\(\Rightarrow x-1\inƯ_{\left(3\right)}=\left(3;-3;1;-1\right)\)

nếu x-1 = 3 => x = 4 (TM)

     x-1  = - 3 => x = -2 (TM)

    x -1  = 1 => x = 2 (TM)

   x -1  = -1 => x = 0 (TM)

KL: x  = ....

b) ta có: \(\frac{x-2}{x+3}=\frac{x+3-5}{x+3}=\frac{x+3}{x+3}-\frac{5}{x+3}=1-\frac{5}{x+3}\)

Để B có giá trị nguyên

\(\Rightarrow\frac{5}{x+3}\in z\)

\(\Rightarrow5⋮x+3\Rightarrow x+3\inƯ_{\left(5\right)}=\left(5;-5;1;-1\right)\)

nếu x+3 = 5 => x = 2(TM)

      x +3 = -5 => x = -8 (TM)

      x +3 = 1 => x = -2 (TM)

    x+3 = -1 => x = -4 (TM)

KL: x =.....

c) ta có: \(\frac{2x+1}{x-3}=\frac{2x-6+7}{x-3}=\frac{2.\left(x-3\right)+7}{x-3}=\frac{2.\left(x-3\right)}{x-3}+\frac{7}{x-3}=2+\frac{7}{x-3}\)

Để C có giá trị nguyên

\(\Rightarrow\frac{7}{x-3}\in z\)

\(\Rightarrow7⋮x-3\Rightarrow x-3\inƯ_{\left(7\right)}=\left(7;-7;1;-1\right)\)

nếu x -3 = 7 => x = 10 (TM)

      x -3 = - 7 => x = -4 (TM)

    x -3 = 1 => x =4 (TM)

   x -3 = -1 => x = 2 (TM)

KL: x =...

d) ta có: \(\frac{x^2-1}{x+1}=\frac{x^2+x-x+1}{x+1}=\frac{x.\left(x+1\right)-(x+1)+1}{x+1}\)

                    \(=\frac{x.\left(x+1\right)}{x+1}-\frac{x+1}{x+1}+\frac{1}{x+1}=x-1+\frac{1}{x+1}\) ( x thuộc Z)

Để D có giá trị nguyên

\(\Rightarrow\frac{1}{x+1}\in z\)

\(\Rightarrow1⋮x+1\Rightarrow x+1\inƯ_{\left(1\right)}=\left(1;-1\right)\)

nếu x +1 = 1 => x =1 (TM)

     x +1 = -1 => x = -2 (TM)

KL: x = ....

Chúc bn học tốt !!!!!!

Hoàn tất đoạn văn sau, sau đó trả lời câu hỏi bên dưới

Quang s camping(1)_______at_____the weekend, he often go camping(2)_____on_____the mountains. He usually goes(3)____with______ his friend. Quang and his fried always wear strong boots(4)_____and______warm clothes. (5)______They_____always take food, water and a camping stove. Sometimes, they (6)______camp_______overnight.

* Questions:

1. What does Quang s?

He s camping.

2. Where does he often go camping?

He often goes camping on the mountains.

3. When does he go?

On weekend.

4. Who does he usually go with?

He usually goes with his friend.

5. What do they always wear?

Quang and his fried always wear strong boots and warm clothes.

6. What do they always take?

They always take food, water and a camping stove.

7. Do they camp overnight?

Yes, they do.

nay, làm gì vậy ta 

ông bị khùng chac