11. David’s school ………….. is very bad this term.

A. report B. period C. day D. semester

12. Learners don't only learn the meaning ........................ the spelling and pronunciation of new words.

A. too B. as well as C. also D. but also

13. He told us to use our dictionaries to ............. every word we didn't understand.

A. look out B. look for C. look at D. look up

14. Hoa is reading short stories …………. Jack London.

A. of B. from C. by D. with

15. John ………….. Susan to go out with him but her parents didn’t let her out.

A. insisted B. suggested C. invited D. helped

16. This book is ……………. that I try to read it from beginning to end at one time.

A. so interesting B. so interested C. such interesting D. too interesting

17. “Let it be” is a famous song ……………. . It is one of the best works by this band.

A. singing by the Beatles B. was sung by the Beatles

C. sung by the Beatles D. sang by the Beatles

18. Surely David's not going to drive, ............. he?

A. does B. isn't C. is D. will

19. She was ………….. of watching television.

A. interested B. tired C. tiring D. pleased

20. My mother ………….. me that I should wake up earlier.

A. told to B. said C. was told D. said to

21. Our teacher asked us ……………. in class.

A. not to talk B. to not talk C. no talk D. without talking

22. We take part ............ most youth activities of our school.

A. of B. on C. in D. at

23. You should …………… your lesson before the examination.

A. read B. look C. revise D. study

24. Hung is …………… do the crossword puzzles.

A. intelligent enough to B. intelligent enough

C. enough intelligent D. enough intelligent to

25. Would you mind if I ............... your car?

A. drove B. to drive C. drive D. driving

26. His father is very proud ……………… him.

A. of B. with C. on D. about

27. Yesterday I came …………. your brother when I was going to school.

A. of B. to C. about D. across

28. Students should work …………. to make their parents happy.

A. hard B. hardly C. in hard way D. more hardly

29. Keep all dangerous things out of children’s …………. .

A. hands B. place C. reach D. head

30. When I was young, I used ……………. on a farm.

A. to live B. live C. to living D. living

IV- Supply the correct form of the word in Capital letters for each gap in the following sentences. Identify your answer by writing the corresponding word or phrase on your answer paper.

31. This coffee is too hot to drink. heat

32. I can’t tell the difference between the twins. differ

33. A fairy appeared and magically changed Miss Tam’s rags into beautiful clothes.magic

34. He drove careless and he had an accident. care

35. I feel so sleepy that I’m going to bed. sleep

36. Why was his lecture so boring ?. bore

37. How many celebration are there in a year in Vietnam ?. celebrate

38. Sciencist have warned the local people of the disaster. science

\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot...\cdot\dfrac{15}{32}=2^x\)

\(\Rightarrow\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot15}{4\cdot6\cdot8\cdot10\cdot...\cdot32}=2^x\Rightarrow\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot15}{\left(2\cdot2\right)\left(2\cdot3\right)\left(2\cdot4\right)\cdot...\cdot\left(2.16\right)}=2^x\)

\(\Rightarrow\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot15}{2^{15}\cdot\left(2\cdot3\cdot4\cdot...\cdot16\right)}=2^x\Rightarrow\dfrac{1}{2^{19}}=2^x\)

\(\Rightarrow2^{-19}=2^x\Rightarrow x=-19\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\left(k\ne0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5.bk+3b}{5.bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)(1)

\(\dfrac{5c+3d}{5c-3d}=\dfrac{5.dk+3d}{5.dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)(2)

Từ (1) và (2) \(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)

a, Xét \(\Delta ABE\) và \(\Delta ACD\) có :

\(\left\{{}\begin{matrix}AB=AC\left(gt\right)\\AD=AE\left(gt\right)\\\widehat{BAC}chung\end{matrix}\right.\) \(\Rightarrow\Delta ABE=\Delta ACD\left(c.g.c\right)\)

=> BE = CD ( 2 cạnh tương ứng )

=> \(\left\{{}\begin{matrix}\widehat{DBK}=\widehat{ECK}\\\widehat{ADK}=\widehat{AEK}\left(1\right)\end{matrix}\right.\) ( 2 góc tương ứng )

b, Ta có : \(\widehat{ADK}+\widehat{BDK}=180^o\Rightarrow\widehat{BDK}=180^o-\widehat{ADK}\) (kề bù) (2)

\(\widehat{AEK}+\widehat{CEK}=180^o\Rightarrow\widehat{CEK}=180^o-\widehat{AEK}\) ( kề bù ) (3)

Từ (1) , (2) và (3) \(\Rightarrow\widehat{BDK}=\widehat{CEK}\)

Lại có : \(\left\{{}\begin{matrix}AB=AC\left(gt\right)\\AD=AE\left(gt\right)\end{matrix}\right.\Rightarrow AB-AD=AC-AE\)

\(\Rightarrow BD=CE\) ( vì ... )

Xét \(\Delta BDK\) và \(\Delta CEK\) có :

\(\left\{{}\begin{matrix}BD=CE\left(cmt\right)\\\widehat{BDK}=\widehat{CEK}\left(cmt\right)\\\widehat{DBK}=\widehat{ECK}\left(cmt\right)\end{matrix}\right.\Rightarrow\Delta KBD=\Delta KCE\left(g.c.g\right)\)

=> BK = CK ( 2 cạnh tương ứng )

c, Xét \(\Delta AKB\) và \(\Delta AKC\) có :

\(\left\{{}\begin{matrix}AKchung\\AB=AC\left(gt\right)\\CK=BK\left(cmt\right)\end{matrix}\right.\Rightarrow\Delta AKB=\Delta AKC\left(c.c.c\right)\)

\(\Rightarrow\widehat{BAK}=\widehat{CAK}\) ( 2 góc tương ứng )

AK nằm giữa hai tia AB và AB

=> AK là tia p/g của \(\widehat{BAC}\)

Vì \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{2x}{3}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{3z}{\dfrac{15}{4}}\)

Áp dụng t/c dãy tỉ số bằng nhau , ta có :

\(\dfrac{2x}{3}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{3z}{\dfrac{15}{4}}=\dfrac{2x+y-3z}{3+\dfrac{4}{3}-\dfrac{15}{4}}=\dfrac{14}{\dfrac{7}{12}}=24\)

\(\Rightarrow\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=24\)

\(\Rightarrow\left\{{}\begin{matrix}2x=72\\3y=96\\4z=120\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=36\\y=32\\z=30\end{matrix}\right.\)

A = 1+ 2 + 2² + 2³ + ... + 2⁹

2A = 2 + 2² + 2³ + 2⁴ + ... + 2⁹

2A - A = ( 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰ ) - ( 1 + 2 + 2² + 2³ +... + 2⁹ )

A = 2¹⁰ - 1

Ta có : B = 5.2⁸ = ( 1 + 4 ) 2⁸ = 2¹⁰ + 2⁸

Vì 2¹⁰ -1 < 2¹⁰ + 2⁸ => A < B

Vậy A < B

Vì \(\left|x+\dfrac{1}{1\cdot2}\right|+\left|x+\dfrac{1}{2\cdot3}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|\ge0\forall x\)

\(\Rightarrow100x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left|x+\dfrac{1}{1\cdot2}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|=x+\dfrac{1}{1\cdot2}+...+x+\dfrac{1}{99\cdot100}\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1\cdot2}+...+\dfrac{1}{99\cdot100}\right)=100x\)

\(\Rightarrow99x+\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)=100x\)

\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}=x\)

\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=x\)

\(\Rightarrow x=1-\dfrac{1}{100}=\dfrac{99}{100}\)

Vì 4x = 5y \(\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}\Rightarrow\dfrac{2x}{10}=\dfrac{y}{4}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{2x}{10}=\dfrac{y}{4}=\dfrac{y-2x}{4-10}=\dfrac{-5}{6}\)

\(\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{-5}{6}\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{25}{6}\\y=\dfrac{-10}{3}\end{matrix}\right.\)

\(\left|x-2\right|-\left|2x+3\right|-x=-2\)

\(\Rightarrow\left|x-2\right|-\left|2x+3\right|=-2+x\)

+) Nếu x \(\ge\) 2 \(\Rightarrow\left\{{}\begin{matrix}x-2\ge0\\2x+3\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-2\right|=x-2\\\left|2x+3\right|=2x+3\end{matrix}\right.\)

\(\Rightarrow x-2-\left(2x+3\right)=-2+x\)

\(\Rightarrow x-2-2x-3-x=-2\)

\(\Rightarrow-2x-5=-2\Rightarrow-2x=3\Rightarrow x=\dfrac{-3}{2}\)( ko t/m x \(\ge\) 2 ) => loại

+) Nếu \(-\dfrac{3}{2}\le x< 2\Rightarrow\left\{{}\begin{matrix}x-2< 0\\2x+3\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-2\right|=-x+2\\\left|2x+3\right|=2x+3\end{matrix}\right.\)

\(\Rightarrow-x+2-\left(2x+3\right)=-2+x\)

\(\Rightarrow-x+2-2x-3-x=-2\)

\(\Rightarrow-4x-1=-2\)

\(\Rightarrow-4x=-1\Rightarrow x=\dfrac{1}{4}\) ( t/m \(-\dfrac{3}{2}\le x< 2\) )

+) Nếu \(x< -\dfrac{3}{2}\Rightarrow\left\{{}\begin{matrix}x-2< 0\\2x+3< 0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x-2\right|=-x+2\\\left|2x+3\right|=-2x-3\end{matrix}\right.\)

\(\Rightarrow-x+2-\left(-2x-3\right)=-2+x\)

\(\Rightarrow-x+2+2x+3-x=-2\)

\(\Rightarrow x-x+5=-2\Rightarrow-2=5\) ( vô lý ) => loại

Vậy x = \(\dfrac{1}{4}\)

\(2x^2+3y^2=77\Rightarrow2\left(x^2+y^2\right)+y^2=77\)

Vì 2( x² + y² ) chẵn ; 77 lẻ

=> y² lẻ (1)

Ta có : \(y^2\ge0\forall y\Rightarrow0\le y^2\le77\)

Kết hợp (1) => y² \(\in\left\{1;9;25;49\right\}\)

=> y \(\in\left\{1;3;5;7\right\}\)

........ (you can continue)