1. successfully

2. urbanized

3. uncomfortable

4. performance

5. dangerous

6. appliances

7. informative

8. unfortunately

9. beautify

10. nationality

1. I have known my English teacher since September

2. This is the most interesting book I have ever read.

3. The boy always spends much time playing games.

4. When Jane was younger, she used to singing a jazz band.

1. Is English taught here?

2. Will she be invited to your wedding party?

3. Has the work been finished by Tom?

4. Were some exercises given by the teacher?

5. Has the window of the laboratory been changed?

6. Is a poem going to be written?

7. Are big cakes being made for the party?

8. Are inquires about the thief being made by the police?

9. Must the test be finished before ten?

10. Will the children be brought home with buses?

11. Has your homework been finished?

Đáp án :

1. D 2. B 3. D 4. B 5. D

6. A 7. A 8. C 9. A 10. A

\(\dfrac{4X-3}{X-1}< 5\Leftrightarrow\dfrac{4X-3}{X-1}-5< 0\Leftrightarrow\dfrac{4X-3-5X+5}{X-1}< 0\)

\(\Leftrightarrow\dfrac{-X+2}{X-1}< 0\Leftrightarrow\left[{}\begin{matrix}-x+2>0\\-x+2< 0\end{matrix}\right.\) và \(\left[{}\begin{matrix}x-1< 0\\x-1>0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x< 2\\x>2\end{matrix}\right.\) và \(\Leftrightarrow\left[{}\begin{matrix}x< 1\\x>1\end{matrix}\right.\)

a) G là trọng tâm của \(\Delta\)ABC nên M

là trung điểm của AC và BG = \(\dfrac{2}{3}BM\)

S_GBC = \(\dfrac{2}{3}s_{MBC}\) (2 tam giác GBC , MBC chung đường cao vẽ từ C đến BM và BG =\(\dfrac{2}{3}BM\))

b) S_MBC = \(\dfrac{1}{2}s_{ABC}\) ( 2 tam giác MBC , ABC chung đường cao vẽ từ B đến AC, MC = \(\dfrac{1}{2}AC\))

Mà S_MBC =\(\dfrac{2}{3}S_{MBC}\) ( câu a ). Do đó S_GBC =\(\dfrac{2}{3}.\dfrac{1}{2}S_{MBC}\) = \(\dfrac{1}{3}S_{ABC}\)

Tương tự có S_GAB = \(\dfrac{1}{3}S_{ABC}\) , S_GAB =\(\dfrac{1}{3}S_{ABC}\)

Tam giác vuông AHB tại H

HM là đường trung tuyến ứng với CH

==> HM = \(\dfrac{AB}{2}\)

==> NP = HM \(\left(\dfrac{AB}{2}\right)\)(1)

Mà NP là đường tb của tam giác ABC

==> NM // BC

==> NM // HP (2)

Từ (1) và (2)

==> MNPH là hình thang cân

Ta có : M\(^2\)= (\(\dfrac{5x-4y}{5x+4y}\))\(^2\) = \(\dfrac{\left(5x-4y\right)^2}{\left(5x+4y\right)^2}\)= \(\dfrac{25x^2+16y^2-40xy}{25x^2+16y^2+40xy}\)

= \(\dfrac{41xy-40xy}{41xy+40xy}=\dfrac{xy}{81xy}=\dfrac{1}{81}=\left(\dfrac{1}{9}\right)^2\)

Mà 4y < 5x < 0 \(\Rightarrow\)5x - 4y > 0 . 5x +4y < 0 \(\Rightarrow\) M < 0

Vậy M = - \(\dfrac{1}{9}\)