a) \(\left(3x-1\right)\left(x+2\right)-\left(x+2\right)^2\)

\(=\left(3x^2+6x-x-2\right)-\left(x+2\right)^2\)

\(=\left(3x^2+5x-2\right)-\left(x^2+4x+4\right)\)

\(=3x^2+5x-2-x^2-4x-4\)

\(=2x^2+x-6\) 

b) \(\left(x-1\right)\left(x+1\right)-\left(x^2-2x+1\right)\)

\(=\left(x^2-1\right)-\left(x^2-2x+1\right)\) 

\(=x^2-1-x^2+2x-1\)

\(=2x-2\)

c) \(\left(x-4\right)\left(4+x\right)+2x\left(x-3\right)\)

\(=\left(x-4\right)\left(x+4\right)+2x\left(x-3\right)\)

\(=\left(x^2-16\right)+2x^2-6x\)

\(=x^2-16+2x^2-6x\)

\(=3x^2-6x-16\)

d) \(\left(x-1\right)\left(x^2-1\right)+\left(x+2\right)^3\)

\(=\left(x^3-x-x^2+1\right)+\left(x^3+6x^2+12x+8\right)\)

\(=x^3-x-x^2+1+x^3+6x^2+12x+8\)

\(=2x^3+5x^2+11x+9\)

e) \(\left(2x-1\right)^2-\left(2x-5\right)\left(x+5\right)\)

\(=\left(4x^2-4x+1\right)-\left(2x^2+10x-5x-25\right)\)

\(=\left(4x^2-4x+1\right)-\left(2x^2+5x-25\right)\)

\(=4x^2-4x+1-2x^2-5x+25\)

\(=2x^2-9x+26\)

f) \(\left(3x+1\right)^2-\left(x^2-1\right)\left(x^2+2\right)\)

\(=\left(9x^2+6x+1\right)-\left(x^4+2x^2-x^2-2\right)\)

\(=\left(9x^2+6x+1\right)-\left(x^4+x^2-2\right)\)

\(=9x^2+6x+1-x^4-x^2+2\)

\(=-x^4+8x^2+6x+3\) 

g) \(\left(x^2+1\right)^2-\left(x^2-1\right)\left(x^2+2\right)\)

\(=\left(x^4+2x^2+1\right)-\left(x^4+2x^2-x^2-2\right)\)

\(=\left(x^4+2x^2+1\right)-\left(x^4+x^2-2\right)\)

\(=x^4+2x^2+1-x^4-x^2+2\)

\(=x^2+3\)

h) \(\left(2x^2-4\right)^2-\left(2x^2+4\right)^2\)

\(=\left(4x^4-16x^2+16\right)-\left(4x^4+16x^2+16\right)\)

\(=4x^4-16x^2+16-4x^4-16x^2-16\)

\(=-32x^2\)

Gọi CTHH của acidic oxide đó là \(X_2O_n\)

\(\%m_X=100\%-56,34\%=43,66\%\)

Ta có : \(\dfrac{x}{2}=\dfrac{M_X}{\%m_X}:\dfrac{M_O}{\%m_O}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{M_X}{43,66\%}:\dfrac{16}{56,34\%}\)

\(\Rightarrow M_X=6,2n\)

Ta có bảng : 

Vậy \(X\) là \(P\) , A là \(P_2O_5\)

Gọi kim loại đó là \(X\) có hóa trị là n 

\(4X+nO_2\underrightarrow{t^o}2X_2O_n\) (1)

\(X_2O_n+2nHCl\rightarrow2XCl_n+nH_2O\) (2)

Ta có : \(n_{X_2O_n}=\dfrac{71,4}{2M_X+16n}\left(mol\right)\)

Theo phương trình (2) \(\Rightarrow n_{XCl_n}=2n_{X_2O_n}=\dfrac{71,4}{M_X+8n}\left(mol\right)\)

\(\Rightarrow m_{XCl_n}=\dfrac{71,4}{M_X+8n}.\left(M_X+35,5n\right)=186,9\)

\(\Rightarrow\dfrac{71,4M_X+2534,7n}{M_X+8n}=186,9\)

\(\Rightarrow71,4M_X+2534,7n=186,9M_X+1495,2n\)

\(\Rightarrow115,5M_X=1039,5n\)

\(\Rightarrow M_X=9n\)

Ta có bảng : 

Vậy X là nhôm ( Al )

Ta có : \(n_{Al_2O_3}=\dfrac{71,4}{102}=0,7\left(mol\right)\)

Theo phương trình (1) \(\Rightarrow n_{Al}=2n_{Al_2O_3}=0,7.2=1,4\left(mol\right)\)

\(\Rightarrow m=m_{Al}=1,4.27=37,8\left(g\right)\)

Từ công thức tinh thể => M có hóa trị II.

a)

Theo đề có:

\(\%_S=\dfrac{32.100\%}{M+96+18n}=14,95\%\)

\(\%_M=\dfrac{M.100\%}{M+96+18n}=29,91\%\)

=> Với `n=3` thì `M=64(Cu)`

=> Công thức tinh thể: \(CuSO_4.3H_2O\)

b)

\(m_{CuSO_4}=\dfrac{85,6}{214}.160=64\left(g\right)\)

Công thức: must + V

I don't think he must go out now.

I don't think he must go out now.

1. you must never come into this room without knocking first

2. I need to buy a small gift for my friend

3. The little boy was very interested in reading picture books

4. She intends to make a book report

5. He seems be very tired after work

6. It is impossible to meet Peter at this time

7. You needn't shout loudly

1. you must never come into this room without knocking first

2. I need to buy a small gift for my friend

3. The little boy was very interested in reading picture books

4. She intends to make a book report

5. He seems to be very tired after work

6. It is impossible to meet Peter at this time

7. You needn't shout loudly

1 Ann facies doing DIY in her free time.

2 My father prefers not eating out.

3 Mr. Smith loves going shopping at weekend.

1 Ann fancies doing DIY in her free time

2 My father prefers not eating out

3 Mr Smith loves going shopping at weekend

   ∆ABC có BE là đường phân giác (gt)

  ∆ABC vuông tại A (gt)

⇒ BC² = AB² + AC² (Pythagore)

⇒ BC² - AB² = AC²

= (3 + 5)²

= 64

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

  Ta có:

BC² = AB² + AC² (Pythagore)

= 6² + 64

= 100

⇒ BC = 10

vì BE là đường phân giác của tam giác ABC nên ta có:

\(\dfrac{AE}{EC}=\dfrac{AB}{BC}=\dfrac{3}{5}\)

\(BC=\dfrac{5}{3}AB\)

áp dụng định lý pythagore vào tam giác ABC ta được:

\(AC^2=AB^2+BC^2\)

tổng độ dài đoạn AC là: 3 + 5 = 8

\(AB^2+BC^2=8^2\\ AB^2+\left(\dfrac{5}{3}AB\right)^2=64\\ AB^2+\dfrac{25}{9}AB^2=64\\ AB^2\cdot\left(1+\dfrac{25}{9}\right)=64\\ AB^2\cdot\dfrac{34}{9}=64\\ AB^2=64:\dfrac{34}{9}=64\cdot\dfrac{9}{34}\\ AB^2=\dfrac{576}{34}\\ AB=\sqrt{\dfrac{576}{34}}\text{≈}4,11\)

độ dài đoạn BC là:

BC² = AC² - AB²

BC² = 64 - 16,8921

BC² = 47,1079

BC = \(\sqrt{47,1079}\) ≈ 6,86

VẬY AB = 4,11; BC  =6,86

Đa thức $2x^4-21x^2+1$ không phân tích thành nhân tử bạn nhé.