1. Ann is very friendly. She lives next door.

Ann who lives next door is very friendly

2. The man is a famous actor. You met him at the party last night.

The man whom you met at the party last night is a famous actor

3. There are some words. They are very difficult to translate.

There are some words which are very difficult to translate

4. The sun provides us with heat and light. It is one of millions of stars in the universe.

The sun, which is one of millions of stars in the universe, provides us with heat and light

5. Students will be punished. Their homework is late.

 Students who homework is late will be punished

6. I was looking for a book this morning. I've found it now.

I've found the book I was looking for this morning

7. Is that the car? You want to buy it.

  Is that the car which you want to buy ?

8. Sandra works in advertising. You were talking to her.

Sandra, who you were talking to, works in advertising.

9. Lake Prespa is a lonely beautiful lake. It‟s on the north Greek border.

Lake Prespa, which is a lonely beautiful lake, is on the north Greek border

10. The little girl ate sweets the whole way. She sat next to me on the coach.

The little girl who sat next to me on the coach ate candy the whole way

1. Ann, who lives next door, is very friendly.

2. The man who/that you met at the party last night is a famous actor.

3. There are some words that/which are very difficult to translate.

4. The sun, which is one of millions of stars in the universe, provides us with heat and light.

5. Students whose homework is late will be punished.

6. I've found the book that/which I was looking for in the morning.

7. Is that the car that/which you want to buy?

8. Sandra, who you were talking to, works in advertising.

9. Lake Prespa, which is on the north Greek border, is a lonely beautiful lake.

10. The little girl who/that sat next to me on the coach ate sweets the whole way.

Tìm min:

Theo BĐT AM-GM thì: $P=a^2+b^2+c^2\ge ab+bc+ac$ hay $P\ge 9$

Vậy $P_{min}=9$. Giá trị này đạt tại $a=b=c=\sqrt{3}$

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Tìm max:

$P=a^2+b^2+c^2=(a+b+c{)}^2-2(ab+bc+ac)=(a+b+c{)}^2-18$

Vì $a,b,c\ge 1$ nên:

$(a-1)(b-1)\ge 0\iff ab+1\ge a+b$

Hoàn toàn tương tự: $bc+1\ge b+c;ac+1\ge a+c$

Cộng lại: $2(a+b+c)\le ab+bc+ac+3=12$

$\Rightarrow a+b+c\le 6$

$\Rightarrow P=(a+b+c{)}^2-18\le 6^2-18=18$

Vậy $P_{max}=18$. Giá trị này đạt tại $(a,b,c)=(1,1,4)$ và hoán vị

Vì 2n + 1 là số chính phương . Mà 2n + 1 là số lẻ

=> 2n + 1 = 1(mod8)

=> n chia hết cho 4

=> n + 1 là số lẻ

=> n + 1 = 1(mod8)

=> n chia hết cho 8

Mặt khác :

3n + 2 = 2(mod3)

=> (n + 1) + (2n + 1) = 2(mod3)

Mà n + 1 và 2n + 1 là các số chính phương lẻ

=> (n + 1) = (2n + 1) = 1(mod3)

=. n chia hết cho 3

Mà (3;8) = 1

Vậy n chia hết cho 24

Vì 2n + 1 là số chính phương . Mà 2n + 1 là số lẻ

=> 2n + 1 = 1(mod8)

=> n chia hết cho 4

=> n + 1 là số lẻ

=> n + 1 = 1(mod8)

=> n chia hết cho 8

Mặt khác :

3n + 2 = 2(mod3)

=> (n + 1) + (2n + 1) = 2(mod3)

Mà n + 1 và 2n + 1 là các số chính phương lẻ

=> (n + 1) = (2n + 1) = 1(mod3)

=. n chia hết cho 3

Mà (3;8) = 1

Vậy n chia hết cho 24

chào bạn gà

\(\left(\frac{1}{a};\frac{1}{b};c\right)=\left(x;y;z\right)\)\(\Rightarrow\)\(x+y+z\le2\)

\(P=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+x+y+z=\Sigma\left(\frac{1}{x^2}+\frac{27}{8}x+\frac{27}{8}x\right)-\frac{23}{4}\left(x+y+z\right)\ge\frac{35}{4}\)

Xét biểu thức phụ : \(\frac{1}{\left(2n+3\right)\sqrt{2n+1}+\left(2n+1\right)\sqrt{2n+3}}=\frac{1}{\sqrt{2n+1}.\sqrt{2n+3}\left(\sqrt{2n+1}+\sqrt{2n+3}\right)}\)

\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{\sqrt{2n+1}.\sqrt{2n+3}\left[\left(2n+3\right)-\left(2n+1\right)\right]}\)

\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{2\sqrt{2n+1}.\sqrt{2n+3}}=\frac{1}{2}\left(\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{2n+3}}\right)\)với \(n\ge1\)

Áp dụng : \(S=\frac{1}{3\sqrt{1}+1\sqrt{3}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+\frac{1}{5\sqrt{7}+7\sqrt{5}}+...+\frac{1}{101\sqrt{103}+103\sqrt{101}}\)

\(=\frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}\right)+...+\frac{1}{2}\left(\frac{1}{\sqrt{101}}-\frac{1}{\sqrt{103}}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}+...+\frac{1}{\sqrt{101}}-\frac{1}{\sqrt{103}}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{103}}\right)\)

DM CHƯA HỌC ĐẾN

axb=2xawfd458uf

\(a^3-a^2b+ab^2-6b^3=0\)

\(\Leftrightarrow\left(a-2b\right)\left(a^2+ab+3b^2\right)=0\left(1\right)\)

Vì a>b>0 =>a²+ab+3b²>0 nên từ (1) ta có a=2b

Vậy biểu thức \(A=\frac{a^4-4b^4}{b^4-4a^4}=\frac{16b^4-4b^4}{b^4-64b^4}=\frac{12b^4}{-63b^4}=-\frac{4}{21}\)