1. In all the world, there (be) ____are______ only 14 mountains that (reach) ____reach______above 8,000 meters.

2. He sometimes (come) ______comes____ to see his parents.

3. When I (come) _____came_____, she (leave) ____had left______for Dalat ten minutes ago.

4. My grandfather never (fly) ____has never flown______ in an airplane, and he has no intention of ever doing so.

5. We have just (decide) ____decided_____ that we (undertake) _____would undertake_______ the job.

6. He told me that he (take) _____would take_____ a trip to California the following week.

7. I knew that this road (be) ____was______ too narrow.

8. Right now I (attend) ______am attending____ class. Yesterday at this time I (attend) __was attending________class.

9. Tomorrow I'm going to leave for home. When I (arrive) __arrive________at the airport, Mary (wait) ____will be waiting______ for me.

10. Margaret was born in 1950. By last year, she (live) __had lived________on this earth for 55 years .

11. The traffic was very heavy. By the time I (get) ______got____to Mary's party, everyone had already (arrive) __arrived________

12. I will graduate in June. I (see) ___will see_______ you in July. By the time I (see) ___see_______ you , I (graduate) ___will have graduate_______.

13. I (visit) _____visited_____ my uncle's home regularly when I (be) ____was______ a child.

14. That book (be) ____has been______ on the table for weeks. Haven't You (not read) ____read______ it yet ?

15. David (wash) ___is washing_______ his hands. He has just (repair) ____repaired______ the TV set.

16. Have You (be) ______been____here before? Yes, I (spend) ____spent______ my holidays here last year.

17. We have never (meet) _____met_____ him. We don't know what he (look) _____looks_____ .

18. The car (be) ____will be______ ready for him by the time he (come) ___comes_______tomorrow.

19. On arriving at home I (find) ____found______that she had just (leave) ____left______a few minutes before.

20. When we (arrive) ____arirve______ in London tonight, it will be probably (rain) __raining________.

đề bài đâu mà trả lời?????????????????????????????????????????????????????????

ahihi

Câu 1: 

a) \(\left(x-\frac{1}{2}\right)^3=-27\) \(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(-3\right)^3\)

\(\Leftrightarrow x-\frac{1}{2}=-3\)\(\Leftrightarrow x=-\frac{5}{2}\)

Vậy \(x=-\frac{5}{2}\)

b( \(2x^2+x=0\)\(\Leftrightarrow x\left(2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\2x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)

Vậy \(x=0\)hoặc \(x=-\frac{1}{2}\)

c) \(5^{x+2}=625\)\(\Leftrightarrow5^{x+2}=5^4\)

\(\Leftrightarrow x+2=4\)\(\Leftrightarrow x=2\)

Vậy \(x=2\)

d) \(\frac{7^x+7^{x+1}+7^{x+2}+7^{x+3}}{2^2.5^2.7^2}=2^2\)

\(\Leftrightarrow7^x+7^{x+1}+7^{x+2}+7^{x+3}=2^2.2^2.5^2.7^2\)

\(\Leftrightarrow7^x+7^x.7+7^x.7^2+7^x.7^3=\left(2.2.5.7\right)^2\)

\(\Leftrightarrow7^x+7^x.7+7^x.49+7^x.343=140^2\)

\(\Leftrightarrow7^x.\left(1+7+49+343\right)=19600\)

\(\Leftrightarrow7^x.400=19600\)

\(\Leftrightarrow7^x=49=7^2\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

Câu 2:

a) \(C=1+4+4^2+4^3+.......+4^{48}\)

\(\Rightarrow4C=4+4^2+4^3+4^4+........+4^{49}\)

\(\Rightarrow4C-C=4^{49}-1\)

\(\Rightarrow3C=4^{49}-1\)

\(\Rightarrow C=\frac{4^{49}-1}{3}\)

b) Ta có: \(3C+1=4^{49}-1+1=4^{49}=4^{7.7}=\left(4^7\right)^7⋮4^7\)( đpcm )

c) \(C=1+4+4^2+4^3+........+4^{48}\)

\(=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+........+\left(4^{46}+4^{47}+4^{48}\right)\)

\(=\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+........+4^{46}.\left(1+4+4^2\right)\)

\(=\left(1+4+4^2\right).\left(1+4^3+....+4^{46}\right)\)

\(=\left(1+4+16\right).\left(1+4^3+........+4^{46}\right)\)

\(=21.\left(1+4^3+.....+4^{46}\right)⋮21\)

Ta có

\(\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=\frac{3y}{60}=\frac{2x}{30}\)

Ap dụng tính chất DTSBN ta có

\(\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}=\frac{2x+3y+z}{30+60+28}=\frac{186}{118}=\frac{93}{59}\)

\(\hept{\begin{cases}\frac{x}{15}=\frac{93}{59}\\\frac{y}{20}=\frac{93}{59}\\\frac{z}{28}=\frac{93}{59}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1395}{59}\\y=\frac{1860}{59}\\z=\frac{2604}{59}\end{cases}}\)

Ta có : \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{3}.\frac{1}{5}=\frac{y}{4}.\frac{1}{5}\Rightarrow\frac{x}{15}=\frac{y}{20}\)

            \(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{5}.\frac{1}{4}=\frac{z}{7}.\frac{1}{4}\Rightarrow\frac{y}{20}=\frac{z}{28}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)

Theo tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=\frac{2x+3y+z}{2.15+3.20+28}=\frac{186}{118}=\frac{93}{59}\)

\(\Rightarrow\frac{x}{15}=\frac{93}{59}\Rightarrow x=\frac{93}{59}.15=\frac{1395}{59}\)

      \(\frac{y}{20}=\frac{93}{59}\Rightarrow y=\frac{93}{59}.20=\frac{1860}{59}\)

       \(\frac{z}{28}=\frac{93}{59}\Rightarrow z=\frac{93}{59}.28=\frac{2604}{59}\)

Vậy : \(\left(x;y;z\right)=\left(\frac{1395}{59};\frac{1860}{59};\frac{2604}{59}\right)\)

Tìm x,y

1+3y/12 =1+5y/5x =1+7y/4x 

Giải:Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}=\frac{1+5y-1-7y}{x}=\frac{-2y}{x}\)

\(\Rightarrow x+3xy=-24y\Rightarrow x+3xy+24y=0\Rightarrow x\left(3y+1\right)+8\left(3y+1\right)=8\)

\(\Rightarrow\left(x+8\right)\left(3y+1\right)=8\)

Đến đây đơn giản rồi.Bạn tự làm nha.....................................

Ta có:1+3y/12=1+5y/5x=1+7y/4z=1+3y+1+7y/12+4x=2+10y​

=> 1+5y/5x=2+10y/12+4x=>2+10y/10x=2+10y/12+4x

=>10x=12+4x

6x=12

=>x=12

bạn thấy x để tìm ý nhé

trình bày giả thiết kết luận đi cậu

chán thế

DỄ VẬY MÀ KO LÀM ĐƯỢC CHỈ CÓ CƯỜI CHO THỐI MŨI THÔI

câu trả lời là:
1:B
2:D
3:C
4:A
5:B