thiếu 

A

A

Câu 31: A

Câu 32: C

Câu 33: D

Câu 34: B

Câu 35: C

32 D

I

1 looked for a job for some months

2 was checked by Mr Brown yesterday

3 the weather was bad, we couldn't go out

4 If it weren't raining heavily, we wouldn't cancel

III

1 It was such a boring film that we left before the end

2 I haven't seen her since 2 years ago

3 They will be punished by the teacher for being so lazy

4 They got up early so that they wouldn't be late for the meeting 

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-\left(2x+1\right)+2x+1-\sqrt[3]{6x+1}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-\dfrac{4x^2}{\sqrt{4x+1}+2x+1}+\dfrac{x^2\left(8x+12\right)}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{6x+1}+\sqrt[3]{\left(6x+1\right)^2}}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\left(-\dfrac{4}{\sqrt{4x+1}+2x+1}+\dfrac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{6x+1}+\sqrt[3]{\left(6x+1\right)^2}}\right)\)

\(=\dfrac{-4}{1+1}+\dfrac{12}{1+1+1}=2\)

Áp dụng hệ thức trong tam giác vuông có:

\(AD=\dfrac{AB}{tan30^0}=4\sqrt{3}\)

\(BD=\dfrac{AB}{sin30^0}=8\)

\(AH^2=\dfrac{AD^2.AB^2}{AD^2+AB^2}=12\Leftrightarrow AH=2\sqrt{3}\)

\(BH=\dfrac{AB^2}{BD}=2\)

Diện tích tam giác ABH: \(S_{ABH}=\dfrac{1}{2}AH.BH=\dfrac{1}{2}.2\sqrt{3}.2=2\sqrt{3}\left(đvdt\right)\)

D

d nha bạn

Không có đáp án nào đúng cả bạn ạ.

32D là đúng rồi á bạn

Các phương trình còn lại đều ra R=0 hoặc là vế trái luôn lớn hơn 0 luôn