b sẽ là snt nhỏ nhất và là 2
còn a là snt gần với số 43 nhứt và là 41

D

A

D

C

B

2/ B
3/ C

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.

Thanks bạn nah

1, A => why did you call

2, A => did

3, D => not to

4, D => had

5, C => before