= (2√2 - 3√2 + 10)√2 - √5

= 2.(√2)2 - 3.(√2)2 + √10.√2 - √5

= 4 - 6 + √20 - √5 = -2 + 2√5 - √5

= -2 + √5

= 0,2.10.√3 + 2|√3 - √5|

s

= 2√3 + 2(√5 - √3)

= 2√3 + 2√5 - 2√3 = 2√5

a: \(=\dfrac{6+4\sqrt{2}}{\sqrt{2}+2+\sqrt{2}}+\dfrac{6-4\sqrt{2}}{\sqrt{2}-2+\sqrt{2}}\)

\(=\dfrac{6+4\sqrt{2}}{2+2\sqrt{2}}+\dfrac{6-4\sqrt{2}}{2\sqrt{2}-2}\)

\(=\dfrac{3+2\sqrt{2}}{\sqrt{2}+1}+\dfrac{3-2\sqrt{2}}{\sqrt{2}-1}\)

=căn 2+1+căn 2-1=2căn 2

b: \(=\dfrac{\sqrt{3}+\sqrt{3+\sqrt{3}}+\sqrt{3}-\sqrt{3+\sqrt{3}}}{1-\sqrt{3}-1}=\dfrac{-2\sqrt{3}}{\sqrt{3}}=-2\)

bạn ơi cho mình hỏi câu b chi tiết hơn đước ko ạ

mình chưa hiểu lắm

1: ĐKXĐ: \(a\ge0\)

\(a,=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\\ b,=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\\ c,=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)

\(=2\left|3-\sqrt{2}\right|+\sqrt{18}-5.1=6-2\sqrt{2}+3\sqrt{2}-5\)

\(=1+\sqrt{2}\)

\(A=2\sqrt{3}+\sqrt{\left(2+\sqrt{3}\right)^2}\\ =2\sqrt{3}+\left|2+\sqrt{3}\right|\\ =2\sqrt{3}+2+\sqrt{3}\\ =3\sqrt{3}+2\)

a, P =  7 + 2 - 51 + 14 2 =  7 + 2 - 7 + 2 = 0

b, Q =  2 3 + 1 - 1 3 - 2 + 6 3 + 3

=  2 3 - 1 2 + 3 + 2 + 6 3 - 3 6

=  4 + 3