part 1

1 which

2 won't

3 because

4 Good idea

5 on

6 therefore

part 2

1D 2F 3A 4B 5C 6E

part 3

1 studied

2 has worked

3 being interviewed

4 to be taken

5 would visit

6 put

7 endangered

8 compulsorily

part 4

1 are -> is

2 me -> to me

3 work -> working

4 when -> since

part 5 

1 written novels for

2 to read bedtime stories

3 to wearing

4 has been absent from 

5 doesn't have a passport

6 were given to Martha

7 play football that

8 whom my teacher was

part 6

1 mankind

2 polluted

3 of

4 naturally

part 7

1F 2F 3T 4T

1: \(2155-\left(174+2155\right)+\left(-68+174\right)\)

=2155-174-2155-68+174

=(2155-2155)+(174-174)-68

=-68

2: \(-25\cdot72+25\cdot21-49\cdot25\)

\(=25\left(-72+21-49\right)\)

\(=25\cdot\left(-100\right)=-2500\)

3: \(35\left(14-23\right)-23\left(14-35\right)\)

\(=35\cdot14-35\cdot23-23\cdot14+23\cdot35\)

\(=35\cdot14-23\cdot14=12\cdot14=168\)

4: \(8154-\left(674+8154\right)+\left(-98+674\right)\)

=8154-674-8154-98+674

=(8154-8154)+(674-674)-98

=0+0-98

=-98

5: \(-25\cdot21+25\cdot72+49\cdot25\)

\(=25\left(-21+72+49\right)\)

\(=25\cdot100=2500\)

6: \(27\left(13-16\right)-16\left(13-27\right)\)

\(=27\cdot13-27\cdot16-16\cdot13+16\cdot27\)

\(=27\cdot13-16\cdot13=11\cdot13=143\)

7: -1911-(1234-1911)

=-1911-1234+1911

=-1234

8: \(156\cdot72+28\cdot156\)

\(=156\cdot\left(72+28\right)\)

\(=156\cdot100=15600\)

9: \(32\cdot\left(-39\right)+16\cdot\left(-22\right)\)

\(=16\cdot\left(-78\right)+16\cdot\left(-22\right)\)

\(=16\cdot\left(-78-22\right)=16\cdot\left(-100\right)=-1600\)

10: -1945-(567-1945)

=-1945-567+1945

=-567

11: \(184\cdot33+67\cdot184\)

\(=184\cdot\left(33+67\right)\)

\(=184\cdot100=18400\)

12: \(44\cdot\left(-36\right)+22\cdot\left(-28\right)\)

\(=22\cdot\left(-72\right)+22\cdot\left(-28\right)\)

\(=22\cdot\left(-72-28\right)=22\cdot\left(-100\right)=-2200\)

a) Ta có: \(P=\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{3\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}+1}\right)\)

\(=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{3\sqrt{a}\left(\sqrt{a}+1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\dfrac{3a+3\sqrt{a}-\left(a-\sqrt{a}+2\sqrt{a}-2\right)}{\sqrt{a}}\)

\(=2+\dfrac{3a+3\sqrt{a}-a+\sqrt{a}-2\sqrt{a}+2}{\sqrt{a}}\)

\(=\dfrac{2\sqrt{a}+2a+2\sqrt{a}+2}{\sqrt{a}}\)

\(=\dfrac{2\left(a+2\sqrt{a}+1\right)}{\sqrt{a}}\)

\(=\dfrac{2\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)

b) Ta có: \(P-6=\dfrac{2\left(\sqrt{a}+1\right)^2-6\sqrt{a}}{\sqrt{a}}\)

\(=\dfrac{2a+4\sqrt{a}+2-6\sqrt{a}}{\sqrt{a}}\)

\(=\dfrac{2\left(a-\sqrt{a}+1\right)}{\sqrt{a}}>0\forall a\) thỏa mãn ĐKXĐ

hay P>6

Bài 1:

A\(\cap\) B=(-3;4]\(\cap\) (2;6)

=(2;4]

A\(\cup\) B=(-3;4]\(\cup\) (2;6)

=(-3;6)

A\B=(-3;4]\(2;6)

=(-3;2]

Bài 2:

a: \(\left(x+2\right)\left(2x^2-3x-2\right)=0\)

=>\(\left(x+2\right)\left(2x^2-4x+x-2\right)=0\)

=>(x+2)(x-2)(2x+1)=0

=>x∈{-2;2;-1/2}

mà x∈Z

nên x∈{2;-2}

=>A={2;-2}

b: -2<=n<=2

mà n là số nguyên

nên n∈{-2;-1;0;1;2}

=>(n+1)∈{-1;0;1;2;3}

=>n(n+1)∈{2;-1;0;2;6}

=>B={-1;0;2;6}

Phần bù của A trong B là \(C_{B}A\) =B\A={-1;0;2;6}\{2;-2}

={-1;0;6}

C 

c

9.      She dislikes talking to strangers.

10.  This camera was bought 2 days ago.

11.  Lan is both clever and beautiful.

12.  Unless you phone her, she will be very sad.

13.  They are independent. (self-reliant/self-sufficient)

14.  Do you mind arranging flowers?