36 on

37 to

38 of

39 off

40 into

41 the

42 the

43 the

44 an

45 -

46 more => as

47 too => either

48 like => as

49 Although => Despite

50 bored => boring

51 has been visited

52 was

53 eating

54 are

Câu 14)

a, Mắc ở bên phải, ngay cạnh chốt dương ( tức là dấu " + " )

b, Do mắc nối tiếp nên

\(I=I_1=I_2=0,54A\) 

c, Ta có

\(U_{13}=U_{12}+U_{23}=4,5+2,5=7V\) 

d, Ko sáng được do 2 đèn mắc nối tiếp nhau

iV .

1 on

2 for

3. of

4.off

5 into

V:

1.the

2.the

3.X

4.an

5.X

VII:

1.visited 

2.was

3.eating

4.is

14:

a: \(A=\dfrac{15x^5y^3-10x^3y^2+20x^4y^4}{5x^2y^2}\)

\(=\dfrac{5x^2y^2\left(3x^3y-10x+4x^2y^2\right)}{5x^2y^2}=3x^3y-10x+4x^2y^2\)

Khi x=-1 và y=2 thì \(A=3\left(-1\right)^3\cdot2-10\cdot\left(-1\right)+4\cdot\left(-1\cdot2\right)^2\)

\(=-6+10+4\cdot4=4+16=20\)

b: \(B=\dfrac{4x^4y^2+3x^4y^3-6x^3y^2}{x^2y^2}=4x^2+3x^2y-6x\)

Khi x=y=-2 thì \(B=4\cdot\left(-2\right)^2+3\cdot\left(-2\right)^2\left(-2\right)-6\left(-2\right)\)

\(=16+12-6\cdot4=28-24=4\)

c: \(C=\dfrac{\dfrac{2}{3}xy\left(-3xy+6-9y^2\right)}{\dfrac{2}{3}xy}=-3xy+6-9y^2\)

Khi x=1/2 và y=4 thì \(C=-3\cdot\dfrac{1}{2}\cdot4+6-9\cdot4^2\)

=-6+6-9*16

=-144

d: \(D=\dfrac{x^2y^2\left(\dfrac{1}{3}y^3-\dfrac{2}{3}x^3\right)}{2x^2y^2}=\dfrac{1}{6}y^3-\dfrac{1}{3}x^3\)

Khi x=-3 và y=3 thì \(D=\dfrac{-1}{3}\left(-3\right)^3-\dfrac{1}{6}\cdot3^3\)

=9-9/2

=9/2

e: \(E=\dfrac{5x^2y\left(4x^3y^3+2xy-y^2\right)}{5x^2y}=4x^3y^3+2xy-y^2\)

Khi x=1 và y=-1 thì \(E=-4-2-1=-7\)

f: G\(=\dfrac{x^2yz\left(7x^3y^3z^2-3x^2z+2y\right)}{x^2yz}=7x^3y^3z^2-3x^2z+2y\)

Khi x=-1;y=1;z=2 thì \(G=7\cdot\left(-1\cdot1\right)^3\cdot2^2-3\cdot\left(-1\right)^2\cdot2+2\cdot1\)

=-7*4-6+2

=-28-4=-32

6. B
7. D
8. C
9. A
10. A
11. A
12. A
13. A
14. B
15. C
16. B
17. C
18. A
19. C
 

a: =>\(2x+7\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

=>\(x\in\left\{-3;-4;-\dfrac{5}{2};-\dfrac{9}{2};-2;-5;-\dfrac{3}{2};-\dfrac{11}{2};-\dfrac{1}{2};-\dfrac{13}{2};\dfrac{5}{2};-\dfrac{19}{2}\right\}\)

b: =>x+2+5 chia hết cho x+2

=>\(x+2\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{-1;-3;3;-7\right\}\)