1 makes

2 doesn't eat

3 goes

4 Do-go

5 Do-agree

6 does-come

7 does-work

8 doesn't -water

9 do-do

10 Do they eat

11 am 

12 rises

13 is

14 doesn't read

15 watches

Ex2

1. you don't go to school everyweekday

Do you go to school every weekday?

2. Lan doesn't play badminton with her friends today

Does Lan  play badminton with her friends today?

3. Hoa doesn't like reading story books

Does Hoa like reading story books

4. Lien and Lan don't in the same class

Do Lien and Lan in the same class

5.  Ba's father doesn't work in a big factory with my father

Does  Ba's father work in a big factory with my father

6 She doesn't study English very well

Does she study English very well

7 They don't tall and thin

Do they tall and thin

8 His mother doesn't cook dinner for the family everyday

Does his mother cook dinner for the family everyday

9 Hoa's mother doesn't go shopping this morning

Ex3

1. Lan usually reads book recess

2 Does Ba go to the market every Sunday

3 Her father often rides his motorbike

4 Nam doesn't have short black hair

5 She usually gets up early

7 was listening - was playing

8 taught

9 playing

10 do you like - like

11 is enjoying

12 rained

13 did Lan go

14 was making

15 are running

16 saw -was having

17 will you be

18 likes

19 is sleeping

20 bought - is

21 were watching

22 will buy

23 am going to buy - am saving

24 is going to rain

25 did you live - lived

26 am having

27 were sleeping

28 is cooking

29 goes - is going

30 boils

31 did you visit

tách ra 10 câu mỗi lần đăng nhé

7 was listening - was playing

8 taught

9 playing

10 do you like - like

11 is enjoying

12 rained

13 did Lan go

14 was making

15 are running

16 saw -was having

17 will you be

18 likes

19 is sleeping

20 bought - is

21 were watching

22 will buy

23 am going to buy - am saving

24 is going to rain

25 did you live - lived

26 am having

27 were sleeping

28 is cooking

29 goes - is going

30 boils

31 did you visit

4: Đặt \(x=\dfrac{a+b}{a-b};y=\dfrac{b+c}{b-c};z=\dfrac{c+a}{c-a}\).

Ta có \(\left(x+1\right)\left(y+1\right)\left(z+1\right)=\dfrac{2a.2b.2c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\left(x-1\right)\left(y-1\right)\left(z-1\right)\)

\(\Rightarrow xy+yz+zx=-1\).

Bất đẳng thức đã cho tương đương:

\(x^2+y^2+z^2\ge2\Leftrightarrow\left(x+y+z\right)^2-2\left(xy+yz+zx\right)-2\ge0\Leftrightarrow\left(x+y+z\right)^2\ge0\) (luôn đúng).

Vậy ta có đpcm

mình xí câu 45,47,51 :>

45. a) Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(\dfrac{1}{a}+\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{4}{2b}\ge\dfrac{\left(1+2\right)^2}{a+2b}=\dfrac{9}{a+2b}\left(đpcm\right)\)

Đẳng thức xảy ra <=> a=b

b) Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{\left(1+1+1\right)^2}{a+b+b}=\dfrac{9}{a+2b}\)(1)

\(\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{b+c+c}=\dfrac{9}{b+2c}\)(2)

\(\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{\left(1+1+1\right)^2}{c+a+a}=\dfrac{9}{c+2a}\)(3)

Cộng (1),(2),(3) theo vế ta có đpcm

Đẳng thức xảy ra <=> a=b=c

47. Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(\dfrac{\left(a+b\right)^2}{c}+\dfrac{\left(b+c\right)^2}{a}+\dfrac{\left(c+a\right)^2}{b}\ge\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+c}=\dfrac{\left[2\left(a+b+c\right)\right]^2}{a+b+c}=\dfrac{4\left(a+b+c\right)^2}{a+b+c}=4\left(a+b+c\right)\)(đpcm)

Đẳng thức xảy ra <=> a=b=c

51 were

52 is going to rain

53 was studying - was chatting

54 does - move

55 were - climbed

56 has - is eating

57 do

58 will come

59 swim

60 has learned - am reading

1 read - cooked 

2 are

3 am going to visit

4 was 

5 ate - was

6 are you going to do - went

45x-38x=1505

=>7x=1505

=>x=215

X*(45-38)=1505

X*7=1505

X=1505:7

X=215

ủa sao m.n toàn giải hộ tui ko vậy,tui bảo nêu cách giải mà 😑

Bài 5:

\(y=m\sqrt{x^2-4x+7}-(3x-4)=\frac{(m^2-9)x^2+(24-4m^2)x+(7m^2-16)}{m\sqrt{x^2-4x+7}+3x-4}\)

Để đths $y$ có TCN thì:\(\lim\limits_{x\to \pm \infty}y\) hữu hạn

Để điều này xảy ra thì $m^2-9=0\Leftrightarrow m=\pm 3$

Kiểm tra lại thấy cả 2 giá trị này đều thỏa mãn. 

Bài 6: Tiệm cận của ĐTHS chứ làm gì có tiệm cận hàm số hả bạn? 

a. 

\(y=\frac{x^2-3x+2}{2x^2+x-1}=\frac{x^2-3x+2}{(2x-1)(x+1)}\)

$(2x-1)(x+1)=0\Leftrightarrow x=\frac{1}{2}$ hoặc $x=-1$

Do đó TCĐ của ĐTHS là $x=\frac{1}{2}$ và $x=-1$

Mặt khác: \(\lim\limits_{x\to \pm \infty}\frac{x^2-3x+2}{2x^2+x-1}=\frac{1}{2}\) nên $y=\frac{1}{2}$ là TCN của ĐTHS.

b.

$x+1=0\Leftrightarrow x=-1$ nên $x=-1$ là TCĐ của đths

$\lim\limits_{x\to \pm \infty}\frac{1-x}{1+x}=-1$ nên $y=-1$ là TCN của đths

6c.

$x+2=0\Leftrightarrow x=-2$ nên $x=-2$ là TCĐ của đths.

\(\lim\limits_{x\to \pm \infty}\frac{2017}{x+2}=0\) nên $y=0$ là TCN của đths.

6d.

\(\lim\limits_{x\to -2+}y=+\infty\) nên $x=-2$ là TCĐ của đths. 

\(\lim\limits_{x\to -\infty}y=\lim\limits_{x\to -\infty}[\frac{1}{x+2}+\frac{4}{x-1-\sqrt{x^2-2x-3}}]=0\) nên $y=0$ là TCN.

Bài 3:

Điện trở tương đương: \(R=\dfrac{U}{I}=\dfrac{30}{0,5}=60\Omega\)

Điện trở R₁: \(R_1=R-R_2=60-20=40\Omega\)

\(I=I_1=I_2=0,5A\left(R_1ntR_2\right)\)

Hiệu điện thế hai đầu mỗi điện trở:

\(U_1=R_1.I_1=40.0,5=20V\)

\(U_2=R_2.I_2=20.0,5=10V\)

Bài 2:

a. Ý nghĩa:

- Điện trở định mức của biến trở con chạy là 100Ω

- Cường độ dòng điện định mức của biến trở con chạy là 2A.

b. HĐT lớn nhất: \(U=R.I=100.2=200V\)

c. Chiều dài dây dẫn: \(R=p\dfrac{l}{S}\Rightarrow l=\dfrac{R.S}{p}=\dfrac{100.2.10^{-6}}{0,5.10^{-6}}=400m\)

1 less crowded - more tranquil

2 faster

3 less easily

4 earlier

5 more well-furnished

\(p=738mmHg=98391,9312Pa\)

Áp suất tại chân cột:\(p_2\)

Áp suất tương ứng với độ cao cột thủy ngân:

\(p=d\cdot h\Rightarrow p=\left(p_2-738\right)\cdot136000Pa\)

Đổi 738 mmHg =0,738 mHg

\(P=d_{Hg}.h=136000\cdot0,738=100368\left(Pa\right)\)