Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người đọc hiểu đề của bạn hơn nhé. 

Người ra đề cho con số tào lao quá, dẫn tới đẳng thức của BĐT này ko xảy ra. (vế trái lớn hơn vế phải tuyệt đối, thậm chí là hơn rất xa).

Ta có:

\(ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow abc\le\dfrac{1}{3\sqrt{3}}\)

\(\Rightarrow8\sqrt{3}abc\le\dfrac{8}{3}\) 

Đặt vế trái BĐT cần chứng minh là P:

\(P=\left[\left(a+1\right)\left(b+1\right)\left(c+1\right)\right]^2+\left[\left(a-1\right)\left(b-1\right)\left(c-1\right)\right]^2\)

\(=\left(abc+a+b+c+ab+bc+ca+1\right)^2+\left(abc+a+b+c-ab-bc-ca-1\right)^2\)

\(=\left(abc+a+b+c+2\right)^2+\left(abc+a+b+c-2\right)^2\)

\(=\left(abc+a+b+c\right)^2+4>\dfrac{8}{3}\ge8\sqrt{3}abc\) (đpcm)

a: Xét ΔAHB vuông tại H có HK là đường cao

nên \(AK\cdot AB=AH^2\left(1\right)\)

Xét ΔAHC vuông tại H có HL là đường cao

nên \(AL\cdot AC=AH^2\left(2\right)\)

Từ (1),(2) suy ra \(AK\cdot AB=AL\cdot AC\)

=>\(\dfrac{AK}{AC}=\dfrac{AL}{AB}\)

Xét ΔAKL và ΔACB có

\(\dfrac{AK}{AC}=\dfrac{AL}{AB}\)

\(\widehat{BAC}\) chung

Do đó: ΔAKL~ΔACB

=>\(\widehat{AKL}=\widehat{ACB}\)

=>\(\widehat{BKL}+\widehat{BCL}=180^0\)

=>BKLC là tứ giác nội tiếp

It was a nice Sunday morning. Minh stayed at home alone and help his parents with the housework. Suddenly, the Sky became very dark. Storm came. Strong winds started to blow with heavy rain. He closed all the windows doors, found himself a shelter in the house. The Sun shone and the birds sang again. Minh was glad. He thought his parents could trust him when they were out for work

b.

\(\left\{{}\begin{matrix}BD\perp AC\\CE\perp AB\end{matrix}\right.\) đồng thời H là giao điểm BD, CE

\(\Rightarrow H\) là trực tâm tam giác ABC

\(\Rightarrow AF\) là đường cao thứ 3 \(\Rightarrow\widehat{AFC}=90^0\)

\(\Rightarrow\widehat{FAC}+\widehat{FCA}=180^0-\widehat{AFC}=90^0\) (1)

ADHE nội tiếp đường tròn đường kính AH, mà M là trung điểm AH

\(\Rightarrow M\) là tâm đường tròn đường kính AH

\(\Rightarrow MA=MD\Rightarrow\Delta MAD\) cân tại M

\(\Rightarrow\widehat{FAC}=\widehat{MDA}\) (2)

D, C cùng thuộc (O) \(\Rightarrow OD=OC\Rightarrow\Delta OCD\) cân tại O

\(\Rightarrow\widehat{FCA}=\widehat{ODC}\) (3)

(1);(2);(3) \(\Rightarrow\widehat{MDA}+\widehat{ODC}=90^0\)

\(\Rightarrow\widehat{MDO}=180^0-\left(\widehat{MDA}+\widehat{ODC}\right)=90^0\)

\(\Rightarrow MD\perp OD\)

\(\Rightarrow MD\) là tiếp tuyến tại D của (O)

\(P=2\left(a+b\right)-ab-7+7=2\left(a+b\right)-ab-\left(a^2+b^2+ab\right)+7\)

\(=2\left(a+b\right)-\left(a^2+2ab+b^2\right)+7\)

\(=2\left(a+b\right)-\left(a+b\right)^2+7\)

\(=8-\left(a+b-1\right)^2\le8\)

\(P_{max}=8\) khi \(\left\{{}\begin{matrix}a+b=1\\a^2+b^2+ab=7\end{matrix}\right.\) \(\Rightarrow\left(a;b\right)=\left(-2;3\right);\left(3;-2\right)\)

3. She speaks English as well as her sister.

4. My birthday is the same as Tom's.

5. My bicycle isn't different from yours.

6. Nam doesn't drive as carefully as her sister does.

8. They haven't lived here as long as us.

9. My father is at the same age as yours.

10. His car doesn't run as fast as a race car.

11. Mr. Phong doesn't teach the same subject as Mr. Ha.

12. My house is at the same height as Nga's.

13. This tree is at the same weight as the tree behind my house.

15. Her sister doesn't sing as beautifully as her.

16. Minh doesn't swim as fast as Phong does.

18. My painting is not as expensive as this painting.

19. This picture is not different from the picture in our room.

20. We don't go out for dinner as much as we used to.

1 I am not as tall as you

2 The last exhibition is not as interestig as this exhibition

3 She speaks English as well as her sister

4 My birthday is the same as Tom's 

5 My bicycle isn't different from yours