a) \(x^2-36=0\)

\(\Leftrightarrow x^2-6^2=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\) 

Vậy: ... 

b) \(x^2-10x+25=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot5+5^2=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

Vậy: ... 

a) \(x^2-36=0\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow x^2=\left(\pm6\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

Vậy \(x\in\left\{6;-6\right\}\)

b) \(x^2-10x+25=0\)

\(\Leftrightarrow x^2-2.x.5+5^2=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

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.

5 to be

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 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