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

= \(\sqrt{\dfrac{4-2\sqrt{3}}{4}}+\dfrac{1-\sqrt{3}}{2}\)

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

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

= 0

b) \(\sqrt{41+6\sqrt{6}-12\sqrt{10}-4\sqrt{15}}+2\sqrt{5}-\sqrt{3}\)

= \(\sqrt{18+20+3+2\sqrt{54}-2\sqrt{360}-2\sqrt{60}}+2\sqrt{5}-\sqrt{3}\)

= \(\sqrt{\left(\sqrt{18}-\sqrt{20}+\sqrt{3}\right)^2}+2\sqrt{5}-\sqrt{3}\)

= \(\sqrt{18}-2\sqrt{5}+\sqrt{3}+2\sqrt{5}-\sqrt{3}\)

= \(\sqrt{18}\)

a: \(=\left(\sqrt{3}-\sqrt{2}+\sqrt{2}\right)\cdot\sqrt{3}=\sqrt{3}\cdot\sqrt{3}=3\)

b: \(=\dfrac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}+\sqrt{41}+2}}\)

\(=\dfrac{8\sqrt{41}}{\sqrt{47+5\sqrt{41}}}\)

\(P=\dfrac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}\)

\(P=\dfrac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}\)

\(P=\dfrac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}=\dfrac{8\sqrt{41}}{2\sqrt{41}}=4\)

*P/S: đã nhỡ làm câu a, câu b bạn Phùng Khánh Linh làm rồi :)

\(P=\dfrac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}=\dfrac{8\sqrt{41}}{\sqrt{41+2.2\sqrt{41}+4}+\sqrt{41-2.2\sqrt{41}+4}}=\dfrac{8\sqrt{41}}{2\sqrt{41}}=4\) \(Q=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\dfrac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{3+2\sqrt{3}+1}}+\dfrac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{3-2\sqrt{3}+1}}=\dfrac{\left(2\sqrt{2}+\sqrt{6}\right)\left(3-\sqrt{3}\right)+\left(2\sqrt{2}-\sqrt{6}\right)\left(3+\sqrt{3}\right)}{9-3}=\dfrac{6\sqrt{2}-2\sqrt{6}+3\sqrt{6}-\sqrt{18}+6\sqrt{2}+2\sqrt{6}-3\sqrt{6}-\sqrt{18}}{6}=\dfrac{12\sqrt{6}-6\sqrt{2}}{6}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)

\(M=\dfrac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}\)

\(M=\dfrac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}\)

\(M=\dfrac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}\)

\(M=\dfrac{8\sqrt{41}}{2\sqrt{41}}=\dfrac{8}{2}=4\)

Vậy M = 4

Học tốt nhé :)

Bạn ơi,đâu có câu thức:a\(^2\)+b\(^2\) đâu?Chỉ có công thức a\(^{2^{ }}\)-b\(^2\) thôi mà?!:)))

b: \(=\left(\sqrt{ab}+\dfrac{2\sqrt{ab}}{a}-\sqrt{\dfrac{a^2+1}{ab}}\right)\cdot\sqrt{ab}\)

\(=ab+\dfrac{2ab}{a}-\sqrt{a^2+1}=ab+2b-\sqrt{a^2+1}\)

c: \(=2\sqrt{6b}-6\sqrt{18}+10\sqrt{12}-\sqrt{48}\)

\(=2\sqrt{6b}-18\sqrt{2}+20\sqrt{3}-4\sqrt{3}\)

\(=2\sqrt{6n}-18\sqrt{2}+16\sqrt{3}\)

d: \(=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{21}}{7}\)

mn ơi giải giúp mik bài não cũng đc a

mình cảm ơn mn nhiều ạ =))

tớ nghĩ tớ giải đc 1-2 bài gì đó nhưng tớ ko bít bấm can lm sao giải cho cậu đc

Bài 1 bạn nhóm , trục như thường nhé :D

Bài 2. \(a.A=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)

\(b.B=\sqrt{17-12\sqrt{2}}-\sqrt{9+4\sqrt{2}}=\sqrt{9-2.2\sqrt{2}.3+8}-\sqrt{8+2.2\sqrt{2}+1}=3-2\sqrt{2}-2\sqrt{2}-1=2-4\sqrt{2}\)

\(c.C=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2.\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{43+30\sqrt{2}}=\sqrt{25+2.3\sqrt{2}.5+18}=5+3\sqrt{2}\)

\(d.D=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)

\(D^2=24-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}=24-2\sqrt{81}=24-18=6\)

\(D=-\sqrt{6}\left(do:D< 0\right)\)

cảm ơn bn nhé!!!

Bài 2: 

a: \(BC=\sqrt{10^2+8^2}=2\sqrt{41}\left(cm\right)\)

\(AH=\dfrac{8\cdot10}{2\sqrt{41}}=\dfrac{40}{\sqrt{41}}\left(cm\right)\)

\(BH=\dfrac{64}{2\sqrt{41}}=\dfrac{32}{\sqrt{41}}\left(cm\right)\)

\(CH=\dfrac{100}{2\sqrt{41}}=\dfrac{50}{\sqrt{41}}\left(cm\right)\)

b: \(\dfrac{AD}{BD}=\dfrac{AH^2}{AB}:\dfrac{BH^2}{AB}=\dfrac{AH^2}{BH^2}\)