làm rồi mà

là sao

N=(x^2-3x-2)(x-1)-x(2x+1)

  = x^3-3x^2-2x-x^2+3x+2-2x^2-x

  =x^3-6x^2+2

  =x^2(x-6)+2

P=(2-5x)3x-(x+2)(x-2)

   = 6x - 15x^2 - x^2  + 2^2

   =6x - 16x^2 + 4

   =2(3x - 8x^2 +2)

N=(x^2-3x-2)(x-1)-x(2x+1)

  = x^3-3x^2-2x-x^2+3x+2-2x^2-x

  =x^3-6x^2+2

  =x^2(x-6)+2

Thay x=-4 vào P ta có 

P=2(-12-128+2)

  =2.-142

  =-284

ok chưa

1. Brush your teeth after meals.

2. Don't play soccer in the street

3. Wash and iron your own clothes.

4. Don't eat too much candy.

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1. She is interested in/ fond of/ keen on playing tennis.

2. No house in the street is older than my house.

3. Although the test was difficult, many students did it well. 

4. It's very important to keep the environment clean.

5. What a wonderful holiday they have!

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1. My sister is a fast runner.

2. Hoang plays football badly.

3. Linh is good at swimming.

4. Peter's favourite sport is badminton.

5. Jack is fatter than Mike.

6. Hugo is interested in playing sports in his free time.

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1. There isn't a bookshelf in my room.

2. Ha is harder than Quang.

3. What a lovely baby!

4. Viết lại bằng từ gì nhỉ?

bài 1:

sắp xếp từ bé đến lớn: 8/12, 15/20, 12/15

Bài 6:

Vì \(m^2+1>0\) nên hs nghịch biến trong khoảng \(\left(-\infty;2m\right)\)

Bài 3:

6: \(x< 0\) nên \(y=\sqrt[3]{x}\) nghịch biến

b.

\(SO\perp\left(ABCD\right)\Rightarrow OM\) là hình chiếu vuông gốc của SM lên (ABCD)

\(\Rightarrow\widehat{SMO}\) là góc giữa SM và (ABCD)

\(BD=a\sqrt{2}\Rightarrow SO=\sqrt{SD^2-OD^2}=\sqrt{3a^2-\left(\dfrac{a\sqrt{2}}{2}\right)^2}=\dfrac{a\sqrt{10}}{2}\)

\(OM=\dfrac{1}{2}AB=\dfrac{a}{2}\)

\(\Rightarrow tan\widehat{SMO}=\sqrt{10}\Rightarrow\widehat{SMO}=...\)

b.

Gọi N là trung điểm AB \(\Rightarrow MN\) là đường trung bình tam giác ABD \(\Rightarrow\left\{{}\begin{matrix}MN=\dfrac{1}{2}BD=\dfrac{a\sqrt{2}}{2}\\MN||BD\end{matrix}\right.\)

\(\Rightarrow\widehat{\left(SM;BD\right)}=\widehat{\left(SM;MN\right)}=\widehat{SMN}\)

\(OM=ON\Rightarrow SN=SM=\sqrt{SO^2+OM^2}=\dfrac{a\sqrt{11}}{2}\)

Định lý hàm cos:

\(cos\widehat{SMN}=\dfrac{SM^2+MN^2-SN^2}{2SM.MN}=\dfrac{\sqrt{22}}{22}\Rightarrow\widehat{SMN}=...\)

Bài 1:

3: ĐKXĐ: x>=1

\(x-\sqrt{x+3+4\sqrt{x-1}}=1\)

=>\(x-\sqrt{x-1+2\cdot\sqrt{x-1}\cdot2+4}=1\)

=>\(x-\sqrt{\left(\sqrt{x-1}+2\right)^2}=1\)

=>\(x-\left|\sqrt{x-1}+2\right|=1\)

=>\(x-\left(\sqrt{x-1}+2\right)=1\)

=>\(x-\sqrt{x-1}-2-1=0\)

=>\(x-1-\sqrt{x-1}-2=0\)

=>\(\left(\sqrt{x-1}\right)^2-2\sqrt{x-1}+\sqrt{x-1}-2=0\)

=>\(\left(\sqrt{x-1}-2\right)\left(\sqrt{x-1}+1\right)=0\)

=>\(\sqrt{x-1}-2=0\)

=>\(\sqrt{x-1}=2\)

=>x-1=4

=>x=5(nhận)

cái gì vậy bạn

? bài ở đâu

ko đăng ảnh đc ạ 

dài thế

Tách rời các bài thì mới có người giải nha 

Bài 3: 

a: \(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{1999}-1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-1998}{1999}\)

\(=\dfrac{1}{1999}\)

b: \(\dfrac{5\cdot18-10\cdot27+15\cdot36}{10\cdot36-20\cdot54+30\cdot72}\)

\(=\dfrac{5\cdot9\left(2-2\cdot3+3\cdot4\right)}{10\cdot18\left(2-2\cdot3+3\cdot4\right)}\)

\(=\dfrac{1}{4}\)