D

A

D

C

B

bài ảo

Câu 6:

a: \(A=\sqrt{8+2\sqrt{10+2\sqrt5}}+\sqrt{8-2\sqrt{10+2\sqrt5}}\)

=>\(A^2=8+2\sqrt{10+2\sqrt5}+8-2\sqrt{10+2\sqrt5}+2\cdot\sqrt{\left(8+2\sqrt{10+2\sqrt5}\right)\left(8-2\sqrt{10+2\sqrt5}\right)}\)

=>\(A^2=16+2\cdot\sqrt{64-4\left(10+2\sqrt5\right)}\)

=>\(A^2=16+2\cdot\sqrt{64-40-8\sqrt5}\)

=>\(A^2=16+2\cdot\sqrt{24-8\sqrt5}\)

=>\(A^2=16+2\cdot\sqrt{4\left(6-2\sqrt5\right)}=16+2\cdot2\cdot\sqrt{\left(\sqrt5-1\right)^2}\)

=>\(A^2=16+4\left(\sqrt5-1\right)=12+4\sqrt5\)

=>\(A=\sqrt{12+4\sqrt5}=\sqrt{2\left(6+2\sqrt5\right)}=\sqrt2\left(\sqrt5+1\right)=\sqrt{10}+\sqrt2\)

b: Đặt \(M=\sqrt{7+\sqrt5}+\sqrt{7-\sqrt5}\)

=>\(M^2=7+\sqrt5+7-\sqrt5+2\cdot\sqrt{49-5}=14+2\cdot\sqrt{44}=14+4\sqrt{11}\)

=>\(M=\sqrt{14+4\sqrt{11}}\)

Ta có: \(B=\frac{\sqrt{7+\sqrt5}+\sqrt{7-\sqrt5}}{\sqrt{7+2\sqrt{11}}}-\sqrt{3-2\sqrt2}\)

\(=\sqrt{\frac{14+4\sqrt{11}}{7+2\sqrt{11}}}-\sqrt{\left(\sqrt2-1\right)^2}=\sqrt2-\left(\sqrt2-1\right)=1\)

c: \(C=\sqrt{\sqrt2+2-2\cdot\sqrt{\sqrt2+1}}+1\)

\(=\sqrt{\sqrt2+1-2\cdot\sqrt{\sqrt2+1}\cdot1+1}+1\)

\(=\sqrt{\left(\sqrt{\sqrt2+1}-1\right)^2}+1=\sqrt{\sqrt2+1}-1+1=\sqrt{\sqrt2+1}\)

I don't know how to play the game.

This department is to the left of the post office.

Phương án 1, nhưng phải sửa là with a ball under the table.

I don't know how to play the game.

This department is to the left of the post office.

Oxit là hợp chất gồm 2 nguyên tố hóa học trong đó có một nguyên tố là oxy

Chắc là Oxit không ăn mòn á , bạn có thể nghiên cứu trên mạng về õit

Đặt A = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\)

\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{6}+\frac{1}{20}+...+\frac{1}{90}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(A=\frac{9}{10}.2\)

\(=\frac{9}{5}\)

Đặt \(A=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\)

=> \(\frac{1}{2}A=\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{45}\right).\frac{1}{2}\)

              \(=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)

             \(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{9.10}\)

              = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{10}\)

           = \(1-\frac{1}{10}=\frac{9}{10}\)

=> A= 9/10:1/2=9/5

1 C

2 B

3 C

4 D

5 B

6 D

7 A

8 C

9 B

10 A

C

I

1A

2D

1. C
2. B
3. C
4. D
5. B
6. D
7. A
8.C
9. B
10. A