13 D

14 A

17 D

21 A

22 D

23 A

28 C

31 C

32 C

Bài 4:

a: k=y/x=7/10

b: y=7/10x

c: Khi x=-6 thì y=-7/10*6=-42/10=-21/5

Khi x=1/7 thì y=1/7*7/10=1/10

c: \(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

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

35 D

36 C

39 B

40 C

44 A

45 B

46 C

47 B

48 D

49 A

50 B

52 D

54 A

55 A

13. D

14. A

17. D

21. A

22. D

23. A

28. C

31. C

32. C

56 If he doesn't change his way of living, he will meet a lot of troubles

57 Doing the homework is important

58 Peter is the most intelligent in this class

59 She is too young to get married

7. How often does he go to the library?
8. ....is the shortest student in his class.
9......is her address?
10... is shorter than Ba.

7 how often does he go to the library

8 nga is the shortest student in the class

9 what is her adress

10 hoa is shorter than ba

em cắt bớt đi

5) BPT \(\Leftrightarrow\dfrac{30x-45}{60}-\dfrac{20x+20}{60}>\dfrac{30}{60}-\dfrac{36-12x}{60}\)

\(\Leftrightarrow10x-65>12x-6\) \(\Leftrightarrow-2x>59\) \(\Leftrightarrow x< -\dfrac{59}{2}\)

  Vậy ...

7) ĐK: \(x\ne-3\)

BPT \(\Leftrightarrow\dfrac{x+1}{x+3}-1>0\) \(\Leftrightarrow\dfrac{x+1}{x+3}-\dfrac{x+3}{x+3}>0\) \(\Leftrightarrow\dfrac{-2}{x+3}>0\)

\(\Rightarrow x+3< 0\) \(\Leftrightarrow x< -3\)  

  Vậy ...

9) Sửa đề: \(\dfrac{x^2+2x+2}{x^2+3}\ge1\)

\(\Leftrightarrow\dfrac{x^2+2x+2}{x^2+3}-1\ge0\) \(\Leftrightarrow\dfrac{x^2+2x+2}{x^2+3}-\dfrac{x^2+3}{x^2+3}\ge0\) \(\Leftrightarrow\dfrac{2x-1}{x^2+3}\ge0\)

\(\Rightarrow2x-1\ge0\) \(\Leftrightarrow x\ge\dfrac{1}{2}\)

  Vậy ...

11) BPT \(\Leftrightarrow3x-2\le0\) \(\Leftrightarrow x\le\dfrac{2}{3}\)

  Vậy ...

10) Ta có: \(\dfrac{2x+1}{x^2+2}\ge1\)

\(\Leftrightarrow\dfrac{2x+1-x^2-2}{x^2+2}\ge0\)

\(\Leftrightarrow-x^2+2x-1\ge0\)

\(\Leftrightarrow x^2-2x+1\le0\)

\(\Leftrightarrow x=1\)

12) Ta có: \(\left(x-2\right)\left(x+1\right)\ge0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x>2\\x< -1\end{matrix}\right.\\\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le-1\end{matrix}\right.\)