Chọn A

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 19:

\(=2\sqrt{a}-\sqrt{(3a)^2}.\sqrt{a}+a\sqrt{a}.\sqrt{16}+\sqrt{\frac{4}{a^4}.36a^5}\)

\(=2\sqrt{a}-3a\sqrt{a}+4a\sqrt{a}+\sqrt{144a}\)

\(=2\sqrt{a}+a\sqrt{a}+\sqrt{144}.\sqrt{a}=2\sqrt{a}+a\sqrt{a}+12\sqrt{a}=14\sqrt{a}+a\sqrt{a}\)

Đáp án A.

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

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

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