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

tk:

vẽ lại mạch ta có RAM//RMN//RNB

đặt theo thứ tự 3 R là a,b,c

ta có a+b+c=1 (1)

điện trở tương đương \(\dfrac{1}{R_{td}}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) \(\Rightarrow I=\dfrac{U}{R_{td}}=9.\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) với a,b,c>0

áp dụng bất đẳng thức cô si cho \(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\)  \(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{\sqrt[3]{abc}}\ge\dfrac{3}{\left(\dfrac{a+b+c}{3}\right)}=\dfrac{9}{a+b+c}=9\)

\(\Leftrightarrow9\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge81\Leftrightarrow I\ge81\) I min =81 ( úi dồi ôi O_o hơi to mà vẫn đúng đá nhỉ)

dấu ''='' xảy ra \(\Leftrightarrow a=b=c\left(2\right)\)

từ (1) (2) \(\Rightarrow a=b=c=\dfrac{1}{3}\left(\Omega\right)\)

vậy ... (V LUN MẤT CẢ BUỔI TỐI R BÀI KHÓ QUÁ EM ĐANG ÔN HSG À )

em ơi chụp cả cái mạch điện a xem nào sao chụp nó bị mất r

Câu 6: B

Câu 7: A

Do vai trò của 3 biến là như nhau, không mất tính tổng quát giả sử \(x>y>z\)

Ta có: \(x-z=\left(x-y\right)+\left(y-z\right)\)

Đặt \(\left\{{}\begin{matrix}x-y=a>0\\y-z=b>0\end{matrix}\right.\)  

Do \(x;z\in\left[0;2\right]\Rightarrow x-z\le2\) hay \(a+b\le2\)

Ta có:

\(P=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2+\dfrac{1}{\left(a+b\right)^2}\ge\dfrac{1}{2}\left(\dfrac{4}{a+b}\right)^2+\dfrac{1}{\left(a+b\right)^2}\)

\(P\ge\dfrac{9}{\left(a+b\right)^2}\ge\dfrac{9}{2^2}=\dfrac{9}{4}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=b\\a+b=2\\\end{matrix}\right.\) \(\Rightarrow a=b=1\) hay \(\left(x;y;z\right)=\left(0;1;2\right)\) và các hoán vị