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TP
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DH
11 tháng 7 2021 lúc 13:39

VII hả e, e chụp thế này sao nhìn rõ đc

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TP
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DH
6 tháng 6 2021 lúc 13:15

III

1 appreciated

2 worried

3 tense

4 confident

5 delighted

6 frustrated

7 calm

8 relaxed

9 self-disciplined

10 depressed

V

1 had resolved

2 has taken

3 Did you say

4 not to take

5 wanted

6 is being repaired

7 taking

8 to think

9 had worked

10 was washing - dropped

11 will find

12 faces

13 turned

14 to give

15 will empathise

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NN
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H24
3 tháng 9 2021 lúc 10:43

Đặt \(A=\dfrac{1}{\sqrt[3]{a+7b}}+\dfrac{1}{\sqrt[3]{b+7c}}+\dfrac{1}{\sqrt[3]{c+7a}}\)

\(A=\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(a+7b\right)}}+\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(b+7c\right)}}+\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(c+7a\right)}}\)

\(\ge\dfrac{4}{\dfrac{8+8+a+7b}{3}}+\dfrac{4}{\dfrac{8+8+b+7c}{3}}+\dfrac{4}{\dfrac{8+8+c+7a}{3}}\ge\dfrac{\left(2+2+2\right)^2}{\dfrac{8+8+a+7b+8+8+b+7c+8+8+c+7a}{3}}\)

\(=\dfrac{36.3}{8\left(a+b+c\right)+48}=\dfrac{3}{2}\)

Vậy \(A_{min}=\dfrac{3}{2}\Leftrightarrow a=b=c=1\)

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VN
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DH
20 tháng 5 2021 lúc 20:58

1 Your sister is interested in doing DIY in her free time, isn't she?

2 Unless that girl study harder, she will fail the exam

3 My father has given up smoking since last year

4 Mary wanted to know what Peter would give his brother on his birthday the day after

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HN
20 tháng 5 2021 lúc 20:58

1. Your sister is interested in doing DIY in her free time, isn't she?

2. Unless that girl studies harder, she'll fail the exam.

3. My father has given up smoking since last year.

4. Mary wanted to know what Peter would give his brother on his birthday the next day.

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DM
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LT
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H24
2 tháng 4 2021 lúc 20:45

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VD
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MP
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H24
30 tháng 4 2021 lúc 16:20

Câu 1: C

Câu 2: B

Câu 3: A

Câu 4: C

Câu 5: B

Câu 6: A

Câu 7: B

Câu 8: A

Câu 9: D

Câu 10: B

Câu 11: B

Câu 12: D

Câu 13: A

Câu 14: B

Câu 15: B

Câu 16: C

Câu 17: C 
Câu 18: \(\left\{{}\begin{matrix}a:D\\b:B\end{matrix}\right.\)

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NT
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TG
14 tháng 7 2021 lúc 19:59

BT A - B hay A.B vậy bn ?

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MY
14 tháng 7 2021 lúc 20:33

\(\)với \(x\ge0,x\ne4\) 

\(=>B=\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right]:\dfrac{3}{\sqrt{x}+1}\)

\(B=\left[\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right].\dfrac{\sqrt{x}+1}{3}\)

\(B=\dfrac{3.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)3}=\dfrac{1}{\sqrt{x}-2}\)

\(=>P=\dfrac{18}{A.B}=\dfrac{18}{\dfrac{x-4}{\sqrt{x}-1}.\dfrac{1}{\sqrt{x}-2}}=\dfrac{18}{\dfrac{\sqrt{x}+2}{\sqrt{x}-1}}\)

\(=\dfrac{18\left(\sqrt{x}-1\right)}{\sqrt{x}+2}\)\(=\dfrac{18\sqrt{x}-18}{\sqrt{x}+2}=\dfrac{18\sqrt{x}}{\sqrt{x}+2}-\dfrac{18}{\sqrt{x}+2}\ge-9\)

dấu"=" xảy ra<=>x=0(tm)

 

 

 

\(\)

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