22B

\(P=\left(sin^2x-1\right)tan^2x+\left(cos^2x-1\right)cot^2x\)

\(=-cos^2x.\dfrac{sin^2x}{cos^2x}+\left(-sin^2x\right)\dfrac{cos^2x}{sin^2x}\)

\(=-sin^2x-cos^2x\)

\(=-\left(sin^2x+cos^2x\right)\)

\(=-1\)

Đáp án C

$CH_3-CH_2-CH_2-CH_2-CH_3 + Cl_2 \to CH_2Cl-CH_2-CH_2-CH_2-CH_3 + HCl$
$CH_3-CH_2-CH_2-CH_2-CH_3 + Cl_2 \to CH_3-CHCl-CH_2-CH_2-CH_3 + HCl$

$CH_3-CH_2-CH_2-CH_2-CH_3 + Cl_2 \to CH_3-CH_2-CHCl-CH_2-CH_3$

21.

\(\left\{{}\begin{matrix}SA\perp AB\\AC\perp AB\end{matrix}\right.\) \(\Rightarrow AB\perp\left(SAC\right)\)

E là trung điểm SA, F là trung điểm SB \(\Rightarrow\) EF là đường trung bình tam giác SAB

\(\Rightarrow EF||AB\Rightarrow EF\perp\left(SAC\right)\)

\(\Rightarrow EF=d\left(F;\left(SEK\right)\right)\)

\(SE=\dfrac{1}{2}SA=\dfrac{3a}{2}\) ; \(EF=\dfrac{1}{2}AB=a\)

 \(SC=\sqrt{SA^2+AC^2}=a\sqrt{13}\Rightarrow SK=\dfrac{2}{3}SC=\dfrac{2a\sqrt{13}}{3}\)

\(\Rightarrow S_{SEK}=\dfrac{1}{2}SE.SK.sin\widehat{ASC}=\dfrac{1}{2}.\dfrac{3a}{2}.\dfrac{2a\sqrt{13}}{3}.\dfrac{2a}{a\sqrt{13}}=a^2\)

\(\Rightarrow V_{S.EFK}=\dfrac{1}{3}EF.S_{SEK}=\dfrac{1}{3}.a.a^2=\dfrac{a^3}{3}\)

\(AB\perp\left(SAC\right)\Rightarrow AB\perp\left(SEK\right)\Rightarrow AB=d\left(B;\left(SEK\right)\right)\)

\(\Rightarrow V_{S.EBK}=\dfrac{1}{3}AB.S_{SEK}=\dfrac{1}{3}.2a.a^2=\dfrac{2a^3}{3}\)

22.

Gọi D là trung điểm AB

Do tam giác ABC đều \(\Rightarrow CD\perp AB\Rightarrow CD\perp\left(SAB\right)\)

\(\Rightarrow CD=d\left(C;\left(SAB\right)\right)\)

\(CD=\dfrac{AB\sqrt{3}}{2}=a\sqrt{3}\) (trung tuyến tam giác đều)

N là trung điểm SC \(\Rightarrow d\left(N;\left(SAB\right)\right)=\dfrac{1}{2}d\left(C;\left(SAB\right)\right)=\dfrac{a\sqrt{3}}{2}\)

\(S_{SAB}=\dfrac{1}{2}SA.AB=a^2\sqrt{3}\) \(\Rightarrow S_{SAM}=\dfrac{1}{2}S_{SAB}=\dfrac{a^2\sqrt{3}}{2}\)

\(\Rightarrow V_{SAMN}=\dfrac{1}{3}.\dfrac{a\sqrt{3}}{2}.\dfrac{a^2\sqrt{3}}{2}=\dfrac{a^3}{4}\)

Lại có:

\(V_{SABC}=\dfrac{1}{3}SA.S_{ABC}=\dfrac{1}{3}.a\sqrt{3}.\dfrac{\left(2a\right)^2\sqrt{3}}{4}=a^3\)

\(\Rightarrow V_{A.BCMN}=V_{SABC}-V_{SANM}=\dfrac{3a^3}{4}\)

Hình vẽ câu 21:

22:

a:

\(\overrightarrow{AD}=2\overrightarrow{DB}\)

=>\(\overrightarrow{AD}=\dfrac{2}{3}\overrightarrow{AB}\)

\(\overrightarrow{CE}=3\overrightarrow{EA}\)

=>\(\overrightarrow{AE}=\dfrac{1}{3}\overrightarrow{EC}\)

=>\(\overrightarrow{AE}=\dfrac{1}{4}\overrightarrow{AC}\)

Xét ΔAED có AM là trung tuyến

nên \(\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AD}+\overrightarrow{AE}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{2}{3}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{8}\overrightarrow{AC}\)

b: \(\overrightarrow{MI}=\overrightarrow{ME}+\overrightarrow{EI}\)

\(=\dfrac{1}{2}\overrightarrow{DE}+\overrightarrow{EC}+\overrightarrow{CI}\)

\(=\dfrac{1}{2}\left(\overrightarrow{DA}+\overrightarrow{AE}\right)+\dfrac{3}{4}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{CB}\)

\(=\dfrac{1}{2}\left(-\dfrac{2}{3}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\right)+\dfrac{3}{4}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{AB}\)

\(=\dfrac{-1}{3}\overrightarrow{AB}+\dfrac{1}{8}\overrightarrow{AC}+\dfrac{3}{4}\overrightarrow{AC}-\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{AB}\)

\(=\dfrac{1}{6}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}\)

22. It's waste of time to try and explain anything to Henry

→ It's not worth the time to try and explain anything to Henry.

23. She didn't like to learn her lessons in the evening

→ They made her  learn her lessons in the evening.

24. Why did you read this book ? It wasn't useful

→ It's wasn't useful.

25. Could you please turn off the TV ?

→ Would you mind turning off the TV?

22.was making

23.was you doing

24.was looking

25.I was looking

26.has been

27.hasn't read

28.have you drunk

29.have heard

30.has lived

20A 21C 22B

Bạn chụp mặt trước nữa thì mình mới giúp được hết nha

B - A - D - C - C