Dự đoán đẳng thức xảy ra tại \(a=b=c=\sqrt{3}\)

Ta có: \(\sqrt{a^2+1}=\sqrt{\frac{1}{4}}.\sqrt{4\left(a^2+1\right)}\le\sqrt{\frac{1}{4}}\left(\frac{4+a^2+1}{2}\right)=\frac{5+a^2}{4}\)

Thiết lập hai bđt còn lại tương tự và cộng theo vế:

\(VP\le3+\frac{1}{2}\left(\frac{15+a^2+b^2+c^2}{2}\right)\)\(=\frac{27+a^2+b^2+c^2}{4}\)

Ta chỉ cần chứng minh: \(ab+bc+ca\ge\frac{27}{4}+\frac{a^2+b^2+c^2}{4}\)

Đến đây chưa nghĩ ra =((

Lạy trời cho con đừng gặp ngõ cụt như nãy nx,làm mà cứ ngõ cụt chán ~v

Lời giải:

\(a+b+c=abc\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\) (do a,b,c dương nên a + b + c  > 0 tức là abc > 0)

Lại có: \(1=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge\frac{9}{ab+bc+ca}\Rightarrow VT=ab+bc+ca\ge9\) (1)

Ta sẽ c/m \(VP=3+\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}\le9\)

\(\Leftrightarrow A=\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}\le6\)

Thật vậy: \(A=\frac{1}{2}\left[\sqrt{4\left(a^2+1\right)}+\sqrt{4\left(b^2+1\right)}+\sqrt{4\left(c^2+1\right)}\right]\)

\(\le\frac{1}{2}\left(\frac{15+a^2+b^2+c^2}{2}\right)=\frac{15+a^2+b^2+c^2}{4}\)

Lại gặp ngõ cụt nữa r,=((Ai đó giúp em vs!!!

\(x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}=4\)

\(\Rightarrow\left(x-2+\frac{1}{x^2}\right)+\left(y^2-2+\frac{1}{y^2}\right)=0\)

\(\Rightarrow\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)

\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2=\left(y-\frac{1}{y}\right)^2=0\)

\(\Leftrightarrow x=y=1=-1\)

Forever Miss You : có cách này nhanh hơn =))

Áp dụng BĐT AM-GM ta có: 

\(x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}\ge2.\sqrt{\frac{x^2.1}{x^2}}+2.\sqrt{\frac{y^2.1}{y^2}}=2+2=4\)

Mà \(x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}=4\)

\(\Rightarrow\hept{\begin{cases}x^2=\frac{1}{x^2}\\y^2=\frac{1}{y^2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x^4=1\\y^4=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\pm1\\y=\pm1\end{cases}}\)

Vậy \(\hept{\begin{cases}x=\pm1\\y=\pm1\end{cases}}\)

híc dài quá mình xin phép tách ra cho đỡ mỏi nhé @@

PASSIVE VOICE ( Câu bị động )

I. Chuyển các câu sau đây thành câu bị động:

1. The gardener has planted some trees. 

-> Some trees have been planted by the gardener.

2. Doctor Brown will give you some advice.

 -> You will be given some advice bt Doctor Brown.

3.A famous designer will redecorate the hotel. 

-> The hotel will be redecorated by a famous designer

4. Someone has broken the crystal vase. 

-> The crystal vase has been broken.

5. His parents have brought him up to be polite. 

-> He has been brought up to be polite by his parents.

6. Fleming discovered penicillin. 

-> Penicillin was discovered by Fleming.

7. They will advertise the product on television.

 -> The product will be advertised on television.

8. Someone is remaking that film. 

-> That film is being remade.

9. Picasso painted that picture. 

-> That picture was painted by Picasso.

10. Did the idea interest you? 

-> Were you interested in the idea?

11. You must leave the bathroom tidy. 

-> The bathroom must be left tidy.

12. You should water this plant daily. 

-> This plant should be watered daily.

13. Our neighbor ought to paint the garage. 

-> The garage ought to be painted by our neighbor.

14. I have to return these books to the library. 

-> These books have to be returned to the library.

15. People know him well. 

-> He is known well.

16. You must dry-clean this shirt. 

-> This shirt must be dry-cleaned.

17. Someone will pay you in ten days.

 -> You will be paid in ten days.

18. You can improve your health with more exercise. 

-> Your health can be improved with more exercise.

19. People must obey the law. 

-> The law must be obeyed.

20.The cleaner is going to clean the kitchen floor. 

-> The kitchen floor is going to be cleaned by the cleaner.

\(\left(2x-3\right)^2=\left(2x-3\right)\left(x-1\right)\)

\(\left(2x-3\right)^2-\left(2x-3\right)\left(x-1\right)=0\)

                             \(\left(2x-3\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-3=0\\x-2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1,5\\x=2\end{cases}}\)

Vay \(x\in\left\{1,5;2\right\}\)

\(\left(2x-3\right)^2=\left(2x-3\right)\left(x-1\right)\)

\(\Leftrightarrow4x^2-9-2x^2+3x-3=0\)

\(\Leftrightarrow2x^2+3x-12=0\)

\(\Leftrightarrow2x^2+3x=12\)

Từ đây bạn làm nốt nhé

Nếu sai thì thông cảm cho mình nha

• Chứng minh: (a² + b²)(c² + d²) ≥ (ac + bd)²
• ↔ (ac)² + (ad)² + (bc)² + (bd)² ≥ (ac)² + 2abcd + (bd)²
• ↔ (ad)² + (bc)² ≥ 2abcd
• ↔ (ad)² - 2abcd + (bc)² ≥ 0
• ↔ (ad - bc)² ≥ 0 luôn đúng
• Dáu "='' khi ad = bc

BĐT Bunhiacopxki:

Áp dụng cho 6 số(1,1,1,a,b,c)

\(\left(1^2+1^2+1^2\right).\left(a^2+b^2+c^2\right)\ge\left(1a+1b+1c\right)^2\)

Chứng minh:

\(\left(ax+by\right)^2\le\left(a^2+b^2\right).\left(x^2+y^2\right)\)

\(\Leftrightarrow a^2x^2+2axby+b^2y^2\le a^2x^2+a^2y^2+b^2x^2+b^2y^2\)

\(\Leftrightarrow2axby\le a^2y^2+b^2x^2\)

\(\Leftrightarrow\left(ay-bx\right)^2\ge0\)( đpcm )