= -37. (12+21-36)

= -37.(-3)

=111

=-12x37-21x37+36x37

=37x(-12-21+36)

=37x3=111

D (chắc thế XD)

d

31 If we had our key, we could get into the house

32 If I hadn't been busy working, I could have seen her off at the airport

33 If the house hadn't been expensive , they wouldn't have bought it

34 If You had listened to what I told you earlier, you wouldn't be asking these questions

35 I wish you turned down the radio

35 I wish I had said goodbye to everybody before I left

37 But for his brother's ingenious plan, the stranded climber wouldn't have been rescue

38 If you hadn't written the letter, we wouldn't have known what happened

25 ko dùng in spite of mà dùng because of nhé

Because of being careless in writing, he got bad marks

26 In spite of being ill, she insisted on going to work

27 In spite of my trying to persuade her, i didn't succeed

Đề yêu cầu điều gì em nhỉ?

ĐKXĐ:

a. \(sinx-1\ne0\Rightarrow sinx\ne1\Rightarrow x\ne\dfrac{\pi}{2}+k2\pi\)

b. \(cos3x+1\ne0\Rightarrow3x\ne-1\Rightarrow3x\ne\pi+k2\pi\)

\(\Rightarrow x\ne\dfrac{\pi}{3}+\dfrac{k2\pi}{3}\)

c. \(\left\{{}\begin{matrix}cos3x\ne0\\2cosx-1\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}3x\ne\dfrac{\pi}{2}+k\pi\\cosx\ne\dfrac{1}{2}\\\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x\ne\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

d. \(cos\left(4x-\dfrac{\pi}{3}\right)\ne0\Rightarrow4x-\dfrac{\pi}{3}\ne\dfrac{\pi}{2}+k\pi\)

\(\Rightarrow x\ne\dfrac{5\pi}{24}+\dfrac{k\pi}{4}\)

????

??????

refer

1.T         2.F

3.T         4.F

5.F         6.F

7.T         8.T

1.He lives in Los Angeles California.

2.His sister's name Emma.

3.He wake up at about half past seven.

4.Yes,he does.

5.He leaves home at a quarter to eight.

6.He has lunch at school.

7.He returns home at a quarter past five after school.

8.He makes his homework at home.

9.Yes,he is.

Câu 5: 

\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)

\(=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)

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

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

\(c5\) \(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)

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

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

\(c4:\Rightarrow\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{GG'}\)

\(\overrightarrow{G'A'}+\overrightarrow{G'B'}+\overrightarrow{G'C'}=\overrightarrow{GG'}\)\(\Leftrightarrow\overrightarrow{G'A}+\overrightarrow{AA'}+\overrightarrow{G'B}+\overrightarrow{B'B}+\overrightarrow{G'C}+\overrightarrow{CC'}=\overrightarrow{GG'}\)

\(\Leftrightarrow\overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}=3\overrightarrow{GG'}\)\(\left(dpcm\right)\)

\(c3:a,\) \(2\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}=2\overrightarrow{IA}+2\overrightarrow{IM}=2\overrightarrow{MI}+2\left(\overrightarrow{IM}+\overrightarrow{MI}\right)=2.\overrightarrow{0}=\overrightarrow{0}\left(đpcm\right)\)

\(b,2\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\)

\(=2\left(\overrightarrow{OI}+\overrightarrow{IA}\right)+\overrightarrow{OI}+\overrightarrow{IB}+\overrightarrow{OI}+\overrightarrow{IC}\)

\(=4\overrightarrow{OI}+2\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}\)\(=4\overrightarrow{OI}+\overrightarrow{0}=4\overrightarrow{OI}\left(đpcm\right)\)

\(c2:\) \(\left\{{}\begin{matrix}3AH=2AB\\3AK=AC\\4BM=3MC\\\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\dfrac{3}{2}\overrightarrow{AH}\\\overrightarrow{AC}=3\overrightarrow{AK}\\\overrightarrow{BM}=\dfrac{3}{7}\overrightarrow{BC}\Rightarrow\overrightarrow{BC}=\dfrac{7}{3}\overrightarrow{BM}\end{matrix}\right.\)

\(\Rightarrow\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=3\overrightarrow{AK}-\dfrac{3}{2}\overrightarrow{AH}\)

\(\Rightarrow\dfrac{7}{3}\overrightarrow{BM}=3\overrightarrow{AK}-\dfrac{3}{2}\overrightarrow{AH}\Leftrightarrow\overrightarrow{BM}=\dfrac{9}{7}\overrightarrow{AK}-\dfrac{9}{14}\overrightarrow{AH}\)