Chọn B

\(\lim\limits_{x\rightarrow+\infty}x^3\left(-1+\dfrac{1}{x}-\dfrac{3}{x^2}+\dfrac{4}{x^3}\right)=+\infty.\left(-1\right)=-\infty\)

Câu 11 :  C

11.

Do \(\lim\limits_{x\rightarrow2^-}\left(1-x^2\right)=1-2^2=-3< 0\)

\(\lim\limits_{x\rightarrow2^-}\left(x-2\right)=0\)

Và: \(x-2< 0\) khi \(x< 2\)

\(\Rightarrow\lim\limits_{x\rightarrow2^-}\dfrac{1-x^2}{x-2}=+\infty\)

11. Linh said she preferred texting to sending letters.

12. My mother said she hoped there wouldn't be traffic jams in 20 years.

13. The teacher asked what a language barrier was.

14. Tom wondered if robots would replace teachers in class. 

15. Jerry said the night before, he had dreamt of flying to Vietnam.

Linh said she preferred texting to sending letters.

My mother said she hoped there wouldn't be traffic jams in 20 years' time.

The teacher asked what a language barrier was.

Tom wondered if robots would replace teachers in class.

Jerry said she had dreamt of flying to Venus the night before.

11. Linh said that she prefered texting to send letter

12. My mother said that she hoped there wouldn't be traffic jams in 20 years time

13. The teacher asked what a language barrier was

14. Tom wondered if robots would replace teachers in classes

15. Jerry said that she hard dreamt of flying to Venus the day before

Theo đinh luật khúc xạ ánh sáng (tại điểm \(I\)) :

\(sini_1=nsinr_1\)

\(\Rightarrow sin45^o=\sqrt{2}\cdot sinr_1\Rightarrow sinr_1=\dfrac{1}{2}\Rightarrow r_1=30^o\)

Tam giác ABC đều\(\Rightarrow\)Góc chiết quang \(\widehat{A}=60^o=r_1+r_2\)

\(\Rightarrow r_2=30^o\)

Xét tại điểm J, theo định luật khúc xạ ánh sáng:

\(sini_2=nsinr_2=\sqrt{2}\cdot sin30^o=\dfrac{\sqrt{2}}{2}\Rightarrow i_2=45^o\)

Góc lệch tia ló ra khỏi lăng kính so với tia tới:

\(D=i_1+i_2-A=45^o+45^o-60^o=30^o\)

Chọn A

Câu 8: A

Câu 9: Tóm tắt:

\(A=300J\)

\(t=1p=60s\)

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\(\text{℘ }=?W\)

Công suất của Nam:

\(\text{℘ }=\dfrac{A}{t}=\dfrac{300}{60}=5\left(W\right)\)