Chọn A

A

Câu 14: C

Câu 18:

Ta có: \(3\sqrt{8a}+\dfrac{1}{4}\sqrt{\dfrac{32a}{25}}-\dfrac{a}{\sqrt{3}}\cdot\sqrt{\dfrac{3}{2a}}-\sqrt{2a}\)

\(=6\sqrt{2a}-\sqrt{2a}+\dfrac{1}{4}\cdot\dfrac{4\sqrt{2a}}{5}-\dfrac{a}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{2a}}\)

\(=5\sqrt{2a}+\dfrac{1}{5}\sqrt{2a}-\dfrac{1}{2}\sqrt{2a}\)

\(=\dfrac{47}{10}\sqrt{2a}\)

Chọn C

Câu 18
\(=3\sqrt{4}.\sqrt{2a}+\frac{1}{4}\sqrt{\frac{16}{25}}.\sqrt{2a}-\sqrt{\frac{a^2}{3}}.\sqrt{\frac{3}{2a}}-\sqrt{2a}\)

\(=6\sqrt{2a}+\frac{1}{5}\sqrt{2a}-\sqrt{\frac{a}{2}}-\sqrt{2a}\)

\(=6\sqrt{2a}+\frac{1}{5}\sqrt{2a}-\sqrt{\frac{1}{4}}.\sqrt{2a}-\sqrt{2a}\)

\(=6\sqrt{2a}+\frac{1}{5}\sqrt{2a}-\frac{1}{2}\sqrt{2a}-\sqrt{2a}=\frac{47}{10}\sqrt{2a}\)

Đáp án C.

Câu 58: B

Câu 59: C

III. 

A. 

1. B

2. C

3. B

4. A

5. D

B. 

1. C

2. B

3. B

4. A

5. C

1 b c b a d

2 c b b a c nha

Đề dài thế này sao giải thích nhanh cho e đc

Part 1

1 C

2 B

3 D

4 C

5 B

6 A

Part 2

1 T

2 F

3 F

4 F

V

1 That old house has just been bought

2 If he doesn't take these pills, he won't be better

3 I suggest taking a train

4 Spending the weekend in the countryside is very wonderful

nhiều thật đấy

Ta có: \(2x^2y\cdot\left(x^3y+4x^3y-4x-x^2y+5x^3y-y\right)\)

\(=2x^2y\cdot\left(10x^3y-4x-x^2y-y\right)\)

\(=20x^5y^2-8x^3y-2x^4y^2-2x^2y^2\)

1 The children were divided into four groups.

2 Is Kuala Lumpur different from Ha Noi?

3 English is a compulsory second language in Malaysia.

4 He was very impressed by the efficiency of the staff.

5 Churches and pagodas are places of worship.

6 I wish I were better at Math.

7 We have been good friends for ages.

8 The official religion in Malaysia is Islam.

9 I haven't seen Sally since my birthday party.

10 She moved to France, but we still keep in touch.

 lỗi hình r ạ

để em viết ra vậy ạ

cho tam giac mnp vuông tại m (mn>mp)  có đường cao mk

a) biết mn=20cm, mp=15cm, tính mk và góc mnp (góc làm tròn đến đơn vị phút).

b) vẽ trung tuyến me của tam giác mnp. từ p vẽ đường thẳng vuông góc với me cắt mn tại d. cm tam giác mnp đồng dạng với tam giác mpd, từ đó suy ra mn.md=np.pk