Lần sau bạn đăng riêng ra nhé

1/

a/  \(\lim\limits_{x\rightarrow2}\dfrac{2^2-4.2-2}{\sqrt{2+7}-5}=3\) 

b/ \(\lim\limits_{x\rightarrow2}\dfrac{2\left(x-2\right)\left(x-\dfrac{1}{2}\right)\left(\sqrt{3x-2}+2\right)}{3\left(x-2\right)}=\dfrac{2\left(2-\dfrac{1}{2}\right)\left(\sqrt{3.2-2}+2\right)}{3}=...\)

c/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{-\dfrac{3x}{x}+\dfrac{1}{x}}{-\sqrt{\dfrac{4x^2}{x^2}-\dfrac{3x}{x^2}+\dfrac{1}{x^2}}+\dfrac{x}{x}}=\dfrac{-3}{-2+1}=3\)

2/ 

\(\lim\limits_{x\rightarrow-1^+}f\left(x\right)=\lim\limits_{x\rightarrow-1^+}\dfrac{2x^2+5x+3}{x+1}=\lim\limits_{x\rightarrow-1^+}\dfrac{2\left(x+1\right)\left(x+\dfrac{3}{2}\right)}{x+1}=1=\lim\limits_{x\rightarrow-1^-}f\left(x\right)\)

\(f\left(x\right)=3.2021+4=...\)

\(\lim\limits_{x\rightarrow1}f\left(x\right)\ne f\left(x\right)\) => ham so gian doan tai x=-1

b/ \(f\left(x\right)=\lim\limits_{x\rightarrow-1}f\left(x\right)\Leftrightarrow3m+4=1\Leftrightarrow m=-1\)

3/ Kia là -ax-b hay cộng nhỉ?

1 send - will receive 

2 do - will improve

3 find - will give

4 will go - has

5 will go- gets

6 doesn't phone - will leave

7 don't study - won't pass

8 rains - won't have to

9 won't be able - watch

10 can't move - isn't

11 study - will pass

Câu 11: B

Câu 12: D

Câu 13: B

Câu 14: A

Câu 15: D

23. Please take the form from your teacher and ask her to sign.

24. Can you show me the nearest way to the post office

25. We are going to plant the flowers in the parks and water them in the afternoon.

26. My brother is very good at repairing household appliances.

đã chụp khó nhìn r mà còn lại thi nữa , combo x2 :'((

Câu 6: B

Câu 7: D

Câu 8: B

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:

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

Em tham khảo thử nha:

Cách mạng tư sản pháp đã trải qua 2 giai đoạn phát triển:

– Giai đoạn 1 : Chế độ quân chủ lập hiến ( 14/7/1789 – 10/8/1792)

– Giai đoạn 2 : Chuyên chính dân chủ cách mạng Gia cô banh (2/6/1793 – 27/7/1794)

D - C 
C - C - C - B - B - A 

DC

CCCBBA 

B

B

C

D

B

D

A

A

B

D

D