\(A=\dfrac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}\cdot\dfrac{1}{\sqrt{3}-\sqrt{2}}\)

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

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

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

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

\(\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}+\sqrt{45-\sqrt{41}}}}:\left(\sqrt{3}-\sqrt{2}\right)\) ( đề)

\(=\frac{8\sqrt{41}}{\sqrt{41}+2-\sqrt{41}-2}:\left(\sqrt{3}-\sqrt{2}\right)\)

\(=2\sqrt{41}:\left(\sqrt{3}-\sqrt{2}\right)\)

\(=2\sqrt{123}+2\sqrt{82}\)

vậy.....................

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

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

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

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

Vậy M = 4

Học tốt nhé :)

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

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

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

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

= \(\sqrt{18}\)

\(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\sqrt{3}+2\sqrt{7}}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}\)

\(=\dfrac{\sqrt{2}}{2}\)

___________

\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

__________

\(\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)

\(=\dfrac{3\cdot2\sqrt{2}-2\cdot2\sqrt{3}+2\sqrt{5}}{3\cdot3\sqrt{2}-2\cdot3\sqrt{3}+3\sqrt{5}}\)

\(=\dfrac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)

\(=\dfrac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}\)

\(=\dfrac{2}{3}\)

a: \(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{2}\)

b: \(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

c: \(=\dfrac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\dfrac{2}{3}\)