Giúp em bài II câu 1 đến câu 3 với ạ
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Những câu hỏi liên quan
LT
1
12 tháng 1 2021
1. illegal
2. traffic jam
3. seatbelt
4. safely
5. railway station
6. safety
7. traffic signs
8. helicopter
9. tricycle
10. boat
HV
19 tháng 11 2021
15 They said they would go to the beach if the weather was nice the day after (loại 1)
16 I told Mike Halen would know what to do if she were late (2)
17 Mark said he would have spoken to Frank if he had seen him the day before (3)
18 The girl said he wouldn't run away if she saw a spider (2)
19 David told her if she had too much to do, she could ask him to help her (2)
20 He told me if I studied harder, I would get high marks in next exam (1)
29 tháng 6 2023
13:
\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}sin\left(\dfrac{pi}{33}\right)\cdot cos\left(\dfrac{pi}{33}\right)\cdot cos\left(\dfrac{2pi}{33}\right)\cdot cos\left(\dfrac{4pi}{33}\right)\cdot cos\left(\dfrac{8pi}{33}\right)\cdot cos\left(\dfrac{16pi}{33}\right)\)
\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{2}\cdot sin\dfrac{2}{33}pi\cdot cos\left(\dfrac{2}{33}pi\right)cos\left(\dfrac{4}{33}pi\right)\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)
\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{2}\cdot sin\dfrac{2}{33}pi\cdot cos\left(\dfrac{2}{33}pi\right)cos\left(\dfrac{4}{33}pi\right)\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{4}\cdot sin\dfrac{4}{33}pi\cdot cos\left(\dfrac{4}{33}pi\right)\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)
\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{8}\cdot sin\dfrac{8}{33}pi\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)
\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{16}\cdot sin\dfrac{16}{33}pi\cdot cos\left(\dfrac{16}{33}pi\right)\)
\(=\dfrac{1}{sin\left(\dfrac{pi}{3}\right)}\cdot\dfrac{1}{32}\cdot sin\dfrac{32}{33}pi\)
=1/32
10:
\(=\dfrac{1}{2}\left[cos100+cos60\right]+\dfrac{1}{2}\cdot\left[cos100+cos20\right]\)
=cos100+1/2*cos20+1/4
6:
sin6*cos12*cos24*cos48
=1/cos6*cos6*sin6*cos12*cos24*cos48
=1/cos6*1/2*sin12*cos12*cos24*cos48
=1/cos6*1/4*sin24*cos24*cos48
=1/cos6*1/8*sin48*cos48
=1/cos6*1/16*sin96
=1/16
HN
0
NT
1
29 tháng 7 2021
Bài 3:
c) Ta có: \(\dfrac{2-x}{5}=\dfrac{x+4}{7}\)
\(\Leftrightarrow14-7x=5x+20\)
\(\Leftrightarrow-7x-5x=20-14\)
\(\Leftrightarrow-12x=6\)
hay \(x=-\dfrac{1}{2}\)
26 tháng 9 2021
Câu 1:
Ta có: \(\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)
\(=6x^2+9x+14x+21-\left(6x^2+33x-10x-55\right)\)
\(=6x^2+23x+21-6x^2-23x+55\)
=76
TD
1
30 tháng 6 2023
8:
\(=\dfrac{cos10-\sqrt{3}\cdot sin10}{sin10\cdot cos10}=\dfrac{2\left(\dfrac{1}{2}\cdot cos10-\dfrac{\sqrt{3}}{2}\cdot sin10\right)}{sin20}=\dfrac{sin\left(30-10\right)}{sin20}=1\)
10:
\(=\left(2-\sqrt{3}\right)^2+\left(2+\sqrt{3}\right)^2\)
=7-4căn 3+7+4căn 3=14
12:
\(=cos^270^0+\dfrac{1}{2}\left[cos60-cos140\right]\)
\(=cos^270^0+\dfrac{1}{2}\cdot\dfrac{1}{2}-\dfrac{1}{2}\cdot2cos^270^0+\dfrac{1}{.2}\)
=1/4+1/2=3/4
27 tháng 6 2021
`D=(sqrt{3}.sqrt{5-2sqrt6})/(sqrt3-sqrt2)-1/(2-sqrt3)`
`=(sqrt3*sqrt{3-2sqrt{3}.sqrt2+2})/(sqrt3-sqrt2)-(2+sqrt3)/(4-3)`
`=(sqrt3.sqrt{(sqrt3-sqrt2)^2})/(sqrt3-sqrt2)-2-sqrt3`
`=sqrt3-2-sqrt3=-2`
H
mn ơi giúp em, nhanh ạ, chỉ cần làm đến bài 2 thôi,bài 1 chỉ cần làm đến câu 17
2
20 tháng 10 2021
Bài 6:
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{41}=\dfrac{b}{29}=\dfrac{c}{30}=\dfrac{a+b}{41+29}=\dfrac{700}{70}=10\)
Do đó: a=410; b=290; c=300
1. are at
2. is wearing
3. standing
4. are drinking
5. are dancing
6. are sitting
7. enjoying
8. chatting
9. go
10. dress
11. travel
12. make
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