1. for => to

2. less => fewer

3. in => on

4. are => is

5. will finish => finish

6. furnitures => furniture

7. for => bỏ

8. between => among

9. was => did

10. many => bỏ

1. I often write for him once every two weeks. 

for => to

2. My father has less days off than Tim's father.

less => fewer

3. Nhung's birthday is in the fifth of September.

in => on

4. People should stop smoking because smoking cigarettes are harmful to their health.

are => is

5.I will wait until you will finish your work.

will finish => finish

6.My father is going to buy some new furnitures for our living room

furnitures => furniture

7. My father bought for me a new dictionary last week.

for me => me

8.Between the three apartments, number 29 is the most suitalbe for you.

between => among

9. Mr. Robison used to be a good swimmer and so was Mr. Brown.

was => did

10. People are living longer, so there are more and more many old people.

many => bỏ

a: Trên cùng một nửa mặt phẳng bờ chứa tia Ox, ta có: \(\widehat{xOy}< \widehat{xOz}\left(30^0< 100^0\right)\)

nên tia Oy nằm giữa hai tia Ox và Oz

=>\(\widehat{xOy}+\widehat{yOz}=\widehat{xOz}\)

=>\(\widehat{yOz}=100^0-30^0=70^0\)

Vì tia Ot nằm trong góc yOz

nên tia Ot nằm giữa hai tia Oy,Oz

=>\(\widehat{yOt}+\widehat{zOt}=\widehat{yOz}\)

=>\(\widehat{zOt}=70^0-20^0=50^0\)

Vì \(\widehat{yOt}< \widehat{zOt}\left(20^0< 50^0\right)\)

nên Ot không là phân giác của góc yOz

b: Vì \(\widehat{zOt}< \widehat{zOx}\left(50^0< 100^0\right)\)

nên tia Ot nằm giữa hai tia Oz và Ox

=>\(\widehat{tOz}+\widehat{tOx}=\widehat{xOz}\)

=>\(\widehat{xOt}=100^0-50^0=50^0\)

Ta có: tia Ot nằm giữa hai tia Ox và Oz

mà \(\widehat{xOt}=\widehat{zOt}\left(=50^0\right)\)

nên Ot là phân giác  của góc xOz

skibidi nhes b

have never been.

I              have never been    (Never/be) to London, but I'd love to go there someday.

Ta có: \(\hat{zOt}+\hat{zOy}=180^0\) (hai góc kề bù)

=>\(\hat{zOt}=180^0-\hat{zOy}\)

ta có: \(\hat{xOt}+\hat{xOy}=180^0\) (hai góc kề bù)

=>\(\hat{xOt}=180^0-\hat{xOy}\)

\(\hat{zOt}-\hat{xOt}\)

\(=\left(180^0-\hat{zOy}\right)-\left(180^0-\hat{xOy}\right)\)

\(=180^0-\hat{zOy}-180^0+\hat{xOy}=\hat{xOy}-\hat{zOy}\)

a: \(\left|x+\dfrac{1}{2}\right|>=0\forall x\)

Dấu '=' xảy ra khi \(x+\dfrac{1}{2}=0\)

=>\(x=-\dfrac{1}{2}\)

b: \(\left|x-\dfrac{1}{3}\right|>=0\forall x\)

=>\(\left|x-\dfrac{1}{3}\right|+2>=2\forall x\)

Dấu '=' xảy ra khi \(x-\dfrac{1}{3}=0\)

=>\(x=\dfrac{1}{3}\)

1. encourages

2. ability

3. repetition

4. attendance

5. yearly

6. reasonable

7. attraction

8. harmful

9. attractive

10. worried

1.My teacher..........encouraged................. me to take the exam (courage)

2. He has the ........ability............to become a professional football player ( able)

3. The .................repetition........ of the question was requested ( repeat )

4. The teacher check for regular ..........attendance............. ( attend )

5. Farmers can protect their land by .........yearly.......... rotation of crops ( year)

6.Well, I think that the prices in Dong Thap supermarket are ........reasonable........

( reason )

7. The ........attraction.......... of the moon for the earth causes tides ( attract)

8. Several toxic gases are very............harmful.............. to our heath ( harm )

9. This company offered a lot of.............attracive..... jobs. ( attract )

10. Minh looks..........worried........ again. What's the matter with him ? ( worry )

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3\left(x-1\right)-4\left(y+3\right)+5\left(z-5\right)}{-3\cdot2-4\cdot4+5\cdot6}\\ =\dfrac{\left(5z-3x-4y\right)+\left(3-12-25\right)}{-6-16+30}=\dfrac{50-34}{8}=2\)

`=>(x-1)/2=2=>x-1=4=>x=5`

`=>(y+3)/4=2=>y+3=8=>y=5` 

`=>(z-5)/6=2=>z-5=12=>z=17` 

"BSĐ: tìm nghiệm nguyên"

`x+2y+xy=5`

`=>x+y(x+2)=5`

`=>(x+2)+y(x+2)=5+2`

`=>(x+2)(y+1)=7`

Ta có bảng: 

Vậy: .. 

nếu (x=1):[1+2y+1/cdot y=5] [1+2y+y=5] [1+3y=5] [3y=4] [y=\frac{4}{3}]

Vậy là (x=1) và (y=\frac{4}{3})

bạn xem có đúng ko nhé

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

\(3\left(2a-b\right)=2\left(a+b\right)\)

\(6a-3b=2a+2b\)

\(4a=5b\)

\(a=\dfrac{5}{4}b\)

Thay vào A ta được:

\(A=\dfrac{\left(\dfrac{5}{4}b\right)^4+5^4}{b^4+4^4}=\dfrac{\dfrac{5^4}{4^4}\left(b^4+4^4\right)}{b^4+4}=\dfrac{5^4}{4^4}\)