9. A
10. D
5. A
6. B
7. D
8. B
 

9 A

10 D

e: \(\dfrac{x^2+3x+9}{x^3+4x^2+4x}\cdot\dfrac{x^2+2x}{x^3-27x}\)

\(=\dfrac{x^2+3x+9}{x\left(x^2+4x+4\right)}\cdot\dfrac{x\left(x+2\right)}{x\left(x^2-27\right)}\)

\(=\dfrac{x^2+3x+9}{\left(x+2\right)^2}\cdot\dfrac{x+2}{x\left(x^2-27\right)}\)

\(=\dfrac{\left(x^2+3x+9\right)}{\left(x+2\right)\cdot x\left(x^2-27\right)}\)

f: \(\dfrac{2x^2+4xy+2y^2}{5x-5y}\cdot\dfrac{15x-15y}{2x^3+2y^3}\)

\(=\dfrac{2\left(x^2+2xy+y^2\right)}{5\left(x-y\right)}\cdot\dfrac{15\left(x-y\right)}{2\left(x^3+y^3\right)}\)

\(=\dfrac{\left(x+y\right)^2}{1}\cdot\dfrac{3}{\left(x+y\right)\left(x^2-xy+y^2\right)}\)

\(=\dfrac{3\left(x+y\right)}{x^2-xy+y^2}\)

g: \(\dfrac{x^3-4x}{x^2-7x+12}\cdot\dfrac{x-4}{x^2-2x}\)

\(=\dfrac{x\left(x^2-4\right)}{\left(x-3\right)\left(x-4\right)}\cdot\dfrac{x-4}{x\left(x-2\right)}\)

\(=\dfrac{x^2-4}{\left(x-3\right)\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x-2\right)}=\dfrac{x+2}{x-3}\)

Em cảm ơn nhìu ạ 😍❤️

4. \(x^2-3x+xy-3y=0\)

\(\Leftrightarrow x\left(x-3\right)+y\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-y\end{matrix}\right.\)

5. \(x^2-8x-3x+24=0\)

\(\Leftrightarrow x\left(x-8\right)-3\left(x-8\right)=0\)

\(\Leftrightarrow\left(x-8\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=3\end{matrix}\right.\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

7. \(3x\left(x-1\right)-x^2+2x-1=0\)

\(\Leftrightarrow3x\left(x-1\right)-\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow3x\left(x-1\right)-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left[3x-\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)

8. \(x^2\left(x-3\right)+18-6x=0\)

\(\Leftrightarrow x^2\left(x-3\right)-6\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\pm\sqrt{6}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[\left(x-5\right)-\left(x-2\right)\right]\left[\left(x-5\right)+\left(x-2\right)\right]=0\)

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

\(\Leftrightarrow-3\left(2x-7\right)=0\)

\(\Leftrightarrow2x-7=0\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

12. \(x^2\left(x-3\right)-4x+12=0\)

\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\)

14. \(3x^2-7x-10=0\)

\(\Leftrightarrow3x^2+3x-10x-10=0\)

\(\Leftrightarrow3x\left(x+1\right)-10\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{10}{3}\end{matrix}\right.\)

#Urushi

4: x^2-3x+xy-3y=0

=>x(x-3)+y(x-3)=0

=>(x-3)(x+y)=0

=>x=3 và x+y=0

=>x=3 và y=-3

6: (x-2)^2-5(2-x)=0

=>(x-2)^2+5(x-2)=0

=>(x-2)(x-2+5)=0

=>(x-2)(x+3)=0

=>x=-3 hoặc x=2

8: x^2(x-3)+18-6x=0

=>x^2(x-3)-6(x-3)=0

=>(x-3)(x^2-6)=0

=>x=3 hoặc \(x=\pm\sqrt{6}\)

10: (x-5)^2-(x-2)^2=0

=>(x-5-x+2)(x-5+x-2)=0

=>-3(2x-7)=0

=>2x-7=0

=>x=7/2

12: x^2(x-3)-4x+12=0

=>x^2(x-3)-4(x-3)=0

=>(x-3)(x^2-4)=0

=>(x-3)(x-2)(x+2)=0

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

14: 3x^2-7x-10=0

=>3x^2-10x+3x-10=0

=>(3x-10)(x+1)=0

=>x=10/3 hoặc x=-1

Question 1: Because the water is polluted, the fish are dead.

Question 2: Because households dump waste into the river, it is polluted.

Question 3: The picture was painted by a local artist.

Question 4: A new house will be rebuilt in the area.

Question 5: If I weren't busy, I could attend your birthday party.

Question 6: If Nigel had satellite TV, he could watch the game.

II/ Write a paragraph (80-100 words)

Learning English effectively requires daily practice and the right methods. First, to learn vocabulary efficiently, you should put new words into different situational sentences and practice writing them daily. Second, reading English stories and books helps widen your vocabulary and retain words much longer. Finally, listening to online English programs every day is an excellent way to imitate native speakers' pronunciation and improve your listening skills. By combining these three habits, you will quickly enhance your overall English proficiency.

D

\(lim\dfrac{2\sqrt{7n^2-2n}}{3n+2}=lim\dfrac{2\sqrt{n^2\left(7-\dfrac{2}{n}\right)}}{3n+2}=lim\dfrac{2n\sqrt{7-\dfrac{2}{n}}}{n\left(3+\dfrac{2}{n}\right)}\)

\(=lim\dfrac{2\sqrt{7-\dfrac{2}{n}}}{3+\dfrac{2}{n}}=\dfrac{2\sqrt{7}}{3}\) \(=\dfrac{a\sqrt{7}}{b}\) 

Suy ra : a/b = 2/3 => a - b = -1 

\(\dfrac{x-1}{x+2}+\dfrac{6x}{x^2-4}=\dfrac{x+1}{2-x}\left(dkxd:x\ne\pm2\right)\)

\(\Leftrightarrow\dfrac{x-1}{x+2}+\dfrac{6x}{\left(x-2\right)\left(x+2\right)}=-\dfrac{x+1}{x-2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)+6x+\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=0\)

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

\(\Leftrightarrow2x^2+6x+4=0\)

\(\Leftrightarrow2x^2+2x+4x+4=0\)

\(\Leftrightarrow2x\left(x+1\right)+4\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x+4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+4=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{-1\right\}\)

Cậu nên tách ra và đăng từng bài nha

III

1 Did you attend the Ban Flower Festival in Dien Bien last year?

2 Is Mua Sap a popular folk dance of the Thai people?

IV

1 They live in Wewbley, in north London

2 Her name is Tracy

3 No, there isn't

4 There are 4 people in his family

V

1  - B

2 - C 

3 - d

4 - f

5 - a

6 - e

7 - g