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nhớ tick nha

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bạn cho đề như thế mình đọc không ra? :D chịu khó chụp rõ hơn hoặc gõ tay đi bạn

Question 7:In your class, there are 54 students and there are 32 boy students.How many girl students are there in your class?Answer: There are 12 girl students. .

Mấy câu kia đâu

Ta có : \(\dfrac{x}{15}\text{=}\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{x}{15}\text{=}\dfrac{-5}{15}\)

\(\Rightarrow x\text{=}-5\)

\(\dfrac{x}{15}\)=\(\dfrac{-1}{3}\)

⇒x.3=15. -1

⇒x.3=-15

⇒x   =-15 ; 3

⇒x   =-5

Vậy x=-5

1. He used ( go) to go fishing when he lived in the countryside.

2. Mr. Minh (work) has worked for his company for 20 years.

3..Hoa‘s teacher asked her (study) to study harder.

4. He’ll try his best ( improve) improving English.

5. I (receive) received a letter from my mother yesterday.

6. My brother hates (listen) listening to pop music.

7. It’s very difficult (reach) to reach a decision.

8. Can you help me(move) move this table.

9. Are you going (take) to take Part in this contest.

10.The Y& Y is planning to help the community by (encourage) encouraging all members to participate in a recycling program.

11.Tom has a toothache. He ought (go) to go to the dentist.

12. She (live) has lived in London since 1980

13. In England the sun (not shine) doesn't shine everyday.

14. The doctor advised him(give up) to give up (smoke) smoking.

15 The Red river (flow) flows to the Gulf of Tonkin.

16. He (start) started working a day ago.

17. You should (practice) practice your English more often.

18. I’m going (travel) to travel to Ha Noi.

19. My mother used to (go) go shopping.

20.They (know) have known each other for some months.

21.Do you like (cook) cooking meals?

22. I (write) have written the letter to Hoa since March

23.Can you (help) help me move this table?
_{\(_{\text{sai cho mk xl}}\)}

lm câu mấy ạ

14.

\(\left(ab+c\right)\left(bc+a\right)\le\dfrac{1}{4}\left(ab+bc+c+a\right)^2=\dfrac{1}{4}\left(a+c\right)^2\left(b+1\right)^2\)

Tương tự:

\(\left(ab+c\right)\left(ca+b\right)\le\dfrac{1}{4}\left(b+c\right)^2\left(a+1\right)^2\)

\(\left(bc+a\right)\left(ca+b\right)\le\dfrac{1}{4}\left(a+b\right)^2\left(c+1\right)^2\)

Nhân vế với vế và khai căn:

\(\left(ab+c\right)\left(bc+a\right)\left(ca+b\right)\le\dfrac{1}{8}\left(a+b\right)\left(b+c\right)\left(c+a\right)\left(a+1\right)\left(b+1\right)\left(c+1\right)\)

Mặt khác ta có:

\(\left(a+1\right)\left(b+1\right)\left(c+1\right)\le\dfrac{1}{27}\left(a+b+c+3\right)^3=8\)

\(\Rightarrow\left(ab+c\right)\left(bc+a\right)\left(ca+b\right)\le\dfrac{1}{8}.8.\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Dấu "=" xảy ra khi...

15.

\(\Leftrightarrow\sum\dfrac{2a^2}{2a^2+\left(b+c-a\right)^2}\le2\)

\(\Leftrightarrow\sum\left(\dfrac{2a^2}{2a^2+\left(b+c-a\right)^2}-1\right)\le2-3\)

\(\Leftrightarrow\sum\dfrac{\left(b+c-a\right)^2}{2a^2+\left(b+c-a\right)^2}\ge1\)

Đặt \(\left\{{}\begin{matrix}b+c-a=x\\c+a-b=y\\a+b-c=z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{y+z}{2}\\b=\dfrac{x+z}{2}\\c=\dfrac{x+y}{2}\end{matrix}\right.\)

\(VT=\sum\dfrac{x^2}{2\left(\dfrac{y+z}{2}\right)^2+x^2}=\sum\dfrac{2x^2}{2x^2+\left(y+z\right)^2}\ge\sum\dfrac{2x^2}{2x^2+2\left(y^2+z^2\right)}=1\)

Dấu "=" xảy ra khi...

GIÚP EM VỚI NHA!!!

\(4hm^24dam^2=40400m^2\\94hm^23dam^2=940300m^2\)

#\(Toru\)

4 \(hm^2\) 4 \(dam^2\) = 4 400 \(m^2\)

94 \(hm^2\) 3 \(dam^2\) = 94 300 \(m^2\)

Đặt \(N=n^4+4n^3+7n^2+6n+3=\left(n^2+n+1\right)\left(n^2+3n+3\right)\)

Do \(n\) và \(n+1\) luôn khác tính chẵn lẻ \(\Rightarrow n^2\) và \(n+1\) khác tính chẵn lẻ

\(\Rightarrow n^2+n+1\) luôn lẻ

Gọi \(d=ƯC\left(n^2+n+1;n^2+3n+3\right)\) \(\Rightarrow d\) lẻ

\(\Rightarrow n^2+3n+3-\left(n^2+n+1\right)⋮d\)

\(\Rightarrow2\left(n+1\right)⋮d\)

\(\Rightarrow n+1⋮d\)

\(\Rightarrow\left(n+1\right)^2⋮d\Rightarrow\left(n+1\right)^2-\left(n^2+n+1\right)⋮d\)

\(\Rightarrow n⋮d\Rightarrow n+1-n⋮d\Rightarrow d=1\)

\(\Rightarrow n^2+n+1\) và \(n^2+3n+3\) nguyên tố cùng nhau

Giả sử tồn tại m nguyên dương thỏa mãn: \(\left(n^2+n+1\right)\left(n^2+3n+3\right)=m^3\)

Hiển nhiên \(m>1\), do \(n^2+n+1\) và \(n^2+3n+3\) nguyên tố cùng nhau, đồng thời \(n^2+3n+3>n^2+n+1\)

\(\Rightarrow\left\{{}\begin{matrix}n^2+n+1=1\\n^2+3n+3=m^3\end{matrix}\right.\)

Từ \(n^2+n+1=1\Rightarrow\left[{}\begin{matrix}n=-1\\n=0\end{matrix}\right.\) đều ko thỏa mãn n nguyên dương

Vậy N luôn luôn ko là lập phương