\(\dfrac{2x+2}{3}< 2+\dfrac{x-2}{2} \Leftrightarrow2\left(2x+2\right)< 12+3\left(x-2\right) \Leftrightarrow4x+4< 3x+6 \Leftrightarrow4x< 3x+2 \Leftrightarrow x< 2\)

3.

Từ BBT ta thấy hàm đồng biến trên các khoảng \(\left(-\infty;-1\right)\) và \(\left(1;+\infty\right)\)

B đúng

4.

Từ BBT ta thấy hàm đồng biến trên các khoảng \(\left(-\infty;-1\right)\) và \(\left(0;1\right)\)

A đúng

1.

B sai (thiếu điều kiện \(f'\left(x\right)=0\) tại hữu hạn điểm)

Câu 2 đề thiếu yêu cầu

Câu 9:

Từ đồ thị ta thấy hàm đồng biến trên các khoảng \(\left(-\infty;0\right)\) và \(\left(2;+\infty\right)\)

\(\Rightarrow\) A đúng do \(\left(-1;0\right)\subset\left(-\infty;0\right)\)

`(1+2cosx)(3-cosx)=0`

\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\dfrac{1}{2}\\cosx=3\left(L\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\pi}{3}+k2\pi\\x=\dfrac{-2\pi}{3}+k2\pi\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{2\pi}{3}+k\pi\)

`(k \in ZZ)`

\(\Leftrightarrow\left[{}\begin{matrix}1+2\cos x=0\\3-\cos x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\cos x=-\dfrac{1}{2}\\\cos x=3\end{matrix}\right.\)

Mà \(-1\le\cos x\le1\)

\(\Rightarrow\cos x=-\dfrac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\pi+k2\pi\\x=\dfrac{4}{3}\pi+k2\pi\end{matrix}\right.\)

Vậy ...

5. 

ĐKXĐ: \(cos\left(x-30^0\right)\ne0\Leftrightarrow x\ne120^0+k180^0\)

Pt tương đương:

\(\left[{}\begin{matrix}tan\left(x-30^0\right)=0\\cos\left(2x-150^0\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-30^0=k180^0\\2x-150^0=90^0+k180^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=30^0+k180^0\\x=120^0+k90^0\end{matrix}\right.\)

Kết hợp ĐKXĐ: \(\Rightarrow x=30^0+k180^0\)

6.

\(\Leftrightarrow2\sqrt{2}sinx.cosx+2cosx=0\)

\(\Leftrightarrow2cosx\left(\sqrt{2}sinx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{5\pi}{4}+k2\pi\end{matrix}\right.\)

5. \(\Leftrightarrow\left[{}\begin{matrix}\tan\left(x-30\right)=0\\\cos\left(2x-150\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-30=360k\\\left[{}\begin{matrix}2x-150=90+360k\\2x-150=270+360k\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=30+360k\\x=120+180k\\x=210+180k\end{matrix}\right.\)

Vậy ...

6, \(\Leftrightarrow2\sqrt{2}\sin x.\cos x+2\cos x=0\)

\(\Leftrightarrow\cos x\left(1+\sqrt{2}\sin x\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\dfrac{3}{2}\pi+k2\pi\end{matrix}\right.\\x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{7}{4}\pi+k2\pi\end{matrix}\right.\)

Vậy ...

Câu 18: B

Câu 19: A

18B

19A

20D

Câu 38: A

Câu 37: C

1. have done

2. has written/ has not finished

3. left/ have never met

4. have you had...?

5. did you do/ did you play

6. bought/ has not worn

7. has taught/ graduated

8. Have you heard...?/ has been/ Have you read...?

9. got/ was/ went

10. earned/ has already spent