M.n giúp em với ạ! Em cần gấp ạ! Em cảm ơn nhìu lắm luôn ạ
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1 am going to send
2 will get
3 will do
4 will be
5 are going to visit
6 will win
7 am going to take
8 are going
9 will go
10 is going to defeat
11 is going to have
12 will never lie
13 will fly
14 won't tell
15 will like
8. As it is raining heavily, we can't go on a picnic
9. I suggest (that) you should get a plumber to check the pipes
10. If you don't succeed, you'll have to try it again
11. If she had enough money, she would buy the dictionary
12. I suggest using gas instead of burning coal
13. I suggest you should install a burglar alarm in your house
14. If we recycle, we will save natural resources
15. If you don't hurry up, you will be late for work
16. If you work too much, you will be tired
17. I suggest putting different kinds of.....
18. Lan broke the glass as she was careless
19. If you litter the place around you, it will be a junk yard
20. Paul suggested using public transportation
21. If she doesn't reduce the use of water and electricity, she will have to.....
22. Since the weather was snowy, we postponed our soccer match
23. Tan suggested that Lam should buy a car
24. Lan suggested that I should look for another job
25. The scientist suggested using public transportation instead of private vehicles
26. giống c12
27. The principle is pleased that the students in grade 9 are good helpers
28. If we put garbage into the bins, we will minimize pollution
29. If you are careless, you will broke the vase
30. If you don't stop cutting trees, I will call the police
31. If you use electricity to catch fish, you will be fined heavily
32. I'm sure that that boy will have a test.....
33. We are disappointed that you threw the rubbish on the street
34. It is extremely important that all of us work together to protect the environment
35. I'm sorry that I couldn't do anything to help you
36. He is sure that environmental pollution in this area can be controlled
37. She was annoyed that he damaged the car yesterday
(a) \(A=\dfrac{3}{x-2}\in Z\)
\(\Rightarrow\left(x-2\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=4\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;0;2;4\right\}.\)
(b) \(B=-\dfrac{11}{2x-3}\in Z\)
\(\Rightarrow\left(2x-3\right)\inƯ\left(11\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=1\\2x-3=-1\\2x-3=11\\2x-3=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=7\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{-4;1;2;7\right\}.\)
(c) \(C=\dfrac{x+3}{x+1}=\dfrac{\left(x+1\right)+2}{x+1}=1+\dfrac{2}{x+1}\in Z\Rightarrow\dfrac{2}{x+1}\in Z\)
\(\Rightarrow\left(x+1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\\x+1=2\\x+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=1\\x=-3\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;-2;0;1\right\}.\)
(d) \(D=\dfrac{2x+10}{x+3}=\dfrac{2\left(x+3\right)+4}{x+3}=2+\dfrac{4}{x+3}\in Z\Rightarrow\dfrac{4}{x+3}\in Z\)
\(\Rightarrow\left(x+3\right)\inƯ\left(4\right)=\left\{\pm1;\pm2\pm4\right\}\)
\(\Rightarrow x\in\left\{-2;-4;-1;-5;1;-7\right\}\)
Câu 1.
Khi mở khóa K:
\(I_m=I_1=0,4A\)
Khi đóng khóa K:
\(I_m=I_1+I_2=0,6\Rightarrow I_2=0,2A\)
\(U_1=0,4\cdot5=2V\)
\(\Rightarrow U_2=U_1=2V\)
\(\Rightarrow U=U_1=U_2=2V\)
\(R_2=\dfrac{U_2}{I_2}=\dfrac{2}{0,2}=10\Omega\)
\(a,=x^2+x+4x+4=\left(x+1\right)\left(x+4\right)\\ b,=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\\ c,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ d,=3\left(x^2-2x+5x-10\right)=3\left(x-2\right)\left(x+5\right)\\ e,=-3x^2+6x-x+2=\left(x-2\right)\left(1-3x\right)\\ f,=x^2-x-6x+6=\left(x-1\right)\left(x-6\right)\\ h,=4\left(x^2-3x-6x+18\right)=4\left(x-3\right)\left(x-6\right)\\ i,=3\left(3x^2-3x-8x+5\right)=3\left(x-1\right)\left(3x-8\right)\\ k,=-\left(2x^2+x+4x+2\right)=-\left(2x+1\right)\left(x+2\right)\\ l,=x^2-2xy-5xy+10y^2=\left(x-2y\right)\left(x-5y\right)\\ m,=x^2-xy-2xy+2y^2=\left(x-y\right)\left(x-2y\right)\\ n,=x^2+xy-3xy-3y^2=\left(x+y\right)\left(x-3y\right)\)
đầu tiên ta tính diện tích hình vuông. lấy 8x8=64(cm2)
sau đó, tính diện tích hình tròn. ta lấy bán kính (8:2=4) nhân với chính nó rồi nhân với 3,14(được 50,24.)
Tiếp theo lấy diện tích hình vuông trừ diện tích hình tròn(64 - 50,24= 13,76). Phần thừa ra có 1 đường chéo cắt qua, nếu có hai đường chéo, phần thừa sẽ đc chia làm 8 phần. diện tích màu xám= 1/8 số đó hay 13,76:8=1,72(cm2).