1/

1. although

2. around

3. where

4. forward

5. and

6. will

2/ 1B 2F 3E 4A 5D 6C

3/

1. daily

2. wearing

3. would do

4. haven't eaten

5. is being carried

6. rises

7. inexperienced

8. equality

4/

1. which -> who

2. pollute -> polluting

5/

1. not to touch that

2. Tom has met is

3. weren't ill

4. may be built

6/

1. environmentalists

2. affect

3. media

4. responsible

7/

1. T

2. T

3. F

4. F

Part 6 

1/ students

2/affect

3/media

4/make

chép đề là sao?

ghi lại đề thì b tự lm dc mà

\(\dfrac{2x-3}{4}.\dfrac{6}{5}=\dfrac{21}{10}\\ \Leftrightarrow\dfrac{2x-3}{4}=\dfrac{7}{4}\\ \Leftrightarrow2x-3=7\\ \Leftrightarrow2x=10\\ \Leftrightarrow x=5\)

x = 5

x = 5

1 C

2 C

3 C

4 C

5 A

6 B

7 A

11 

12 B

13 A

14 C

15D

4. B

5. A

6. C

7. C

8. B

9. B

10. A

11. C

12. 

13. A

14. 

15. C

Lời giải:

a.

\(G=\frac{x^2-4}{x+1}+\frac{2}{x+1}:\frac{(2x-3)(x+1)-(2x+1)(x-1)}{(x-1)(x+1)}\)

\(=\frac{x^2-4}{x+1}+\frac{2}{x+1}:\frac{-2}{(x-1)(x+1)}=\frac{x^2-4}{x+1}+\frac{2}{x+1}.\frac{(x+1)(x-1)}{-2}\)

\(=\frac{x^2-4}{x+1}-(x-1)=\frac{x^2-4-(x^2-1)}{x+1}=\frac{-3}{x+1}\)

b.

Để $A\in\mathbb{Z}^+$ thì $x+1$ là ước âm của $-3$

$\Rightarrow x+1\in\left\{-1;-3\right\}$

$\Leftrightarrow x\in\left\{-2;-4\right\}$ (tm)

c.

$G< -1\Leftrightarrow \frac{-3}{x+1}+1< 0$

$\Leftrightarrow \frac{x-2}{x+1}< 0$

$\Leftrightarrow x-2<0< x+1$ hoặc $x-2>0>x+1$

$\Leftrightarrow -1< x< 2$ (chọn) hoặc $-1> x>2$ (loại)

Vậy $-1< x< 2$ và $x\neq 1$

Bài 8:

a: Ta có: \(G=\dfrac{x^2-4}{x+1}+\dfrac{2}{x+1}:\left(\dfrac{2x-3}{x-1}-\dfrac{2x+1}{x+1}\right)\)

\(=\dfrac{x^2-4}{x+1}+\dfrac{2}{x+1}:\dfrac{2x^2+2x-3x-3-2x^2+2x-x+1}{\left(x-1\right)\left(x+1\right)}\)

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

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

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

\(=\dfrac{x^2-4-x^2+1}{x+1}\)

\(=-\dfrac{3}{x+1}\)

c: Để G<-1 thì G+1<0

\(\Leftrightarrow\dfrac{-3+x+1}{x+1}< 0\)

\(\Leftrightarrow\dfrac{x-2}{x+1}< 0\)

\(\Leftrightarrow-1< x\le2\)

Kết hợp ĐKXĐ, ta được:

\(\left\{{}\begin{matrix}-1< x\le2\\x\ne1\end{matrix}\right.\)

a) \(D=\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)\)\(=\left(\dfrac{2}{x+2}-\dfrac{4}{\left(x+2\right)^2}\right):\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right)\)

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

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

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

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

\(=\dfrac{-2.\left(x-2\right)}{x+2}\)

\(x^2-5x+6=0\\ \Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\\ \Rightarrow\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

\(P=\dfrac{-2.\left(x-2\right)}{x+2}\)

Thay \(x=2\), ta có:

\(P=\dfrac{-2.\left(2-2\right)}{2+2}\)

    \(=0\)

Thay \(x=3\), ta có:

\(P=\dfrac{-2.\left(3-2\right)}{3+2}\)

    \(=-\dfrac{2}{5}\)

D nguyên âm \(\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\left(x-2\right)< 0\\x+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}-2\left(x-2\right)>0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -2\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x>2\\x< -2\end{matrix}\right.\)

a:Ta có: \(D=\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)\)

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

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

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

b: Ta có: \(x^2-5x+6=0\)

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

hay x=3

Thay x=3 vào D, ta được:

\(D=\dfrac{-\left(3-2\right)}{3+2}=-\dfrac{1}{5}\)