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

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

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

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

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

\(1.\sqrt{\left(5+\sqrt{7}\right)^2}-\sqrt{8-2\sqrt{7}}=5+\sqrt{7}-\sqrt{7-2\sqrt{7}+1}=5+\sqrt{7}-\sqrt{7}+1=6\)

\(2.\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-2\sqrt{3}}=\sqrt{3}+1-\sqrt{3-2\sqrt{3}+1}=\sqrt{3}+1-\sqrt{3}+1=2\)

\(3.VT=\sqrt{11}-\sqrt{20-6\sqrt{11}}=\sqrt{11}-\sqrt{11-2.3\sqrt{11}+9}=\sqrt{11}-\sqrt{11}+3=3=VP\)

Vậy , đẳng thức được chứng minh .

\(4.VT=\sqrt{41+12\sqrt{5}}-\sqrt{41-12\sqrt{5}}=\sqrt{36+2.6\sqrt{5}+5}-\sqrt{36-2.6\sqrt{5}+5}=6+\sqrt{5}-6+\sqrt{5}=2\sqrt{5}=VP\)

Vậy , đẳng thức được chứng minh .

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

\(\sqrt{\left(\sqrt{10}-3\right)^2}+\sqrt{\left(\sqrt{10}-4\right)^2}=\left|\sqrt{10}-3\right|+\left|\sqrt{10}-4\right|\)

\(=\sqrt{10}-3+4-\sqrt{10}=1\)

\(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}=\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)

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

      \(\sqrt{41-12\sqrt{5}}-\sqrt{41+12\sqrt{5}}=\sqrt{\left(6-\sqrt{5}\right)^2}-\sqrt{\left(6+\sqrt{5}\right)^2}\)

\(=6-\sqrt{5}-6-\sqrt{5}=-2\sqrt{5}\)

   \(A=\sqrt{49a^2}+3a=7\left|a\right|+3a\)

Nếu  \(a\ge0\)thì:   \(A=7a+3a=10a\)

Nếu  \(a< 0\)thì:  \(A=-7a+3a=-4a\)

   \(B=3\sqrt{9a^6}-6a^3=9\left|a^3\right|-6a^3\)

Nếu  \(a\ge0\)thì:  \(B=9a^3-6a^3=3a^3\)

Nếu  \(a< 0\)thì:  \(B=-9a^3-6a^3=-15a^3\)

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

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

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

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

= \(\sqrt{18}\)

\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)

\(\sqrt{8-2\sqrt{12}}=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}=\left|\sqrt{6}-\sqrt{2}\right|=\sqrt{6}-\sqrt{2}\)

\(\sqrt{21+6\sqrt{6}}=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}=\left|3\sqrt{2}-\sqrt{3}\right|=3\sqrt{2}-\sqrt{3}\)

\(\sqrt{15-6\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}=\left|3-\sqrt{6}\right|=3-\sqrt{6}\)

\(\sqrt{29-12\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}=\left|2\sqrt{5}-3\right|=2\sqrt{5}-3\)

\(\sqrt{41+12\sqrt{5}}=\sqrt{\left(6+\sqrt{5}\right)^2}=6+\sqrt{5}\)

C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}}{\sqrt{10}-\sqrt{2}}\)

C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{\left(\sqrt{20}+1\right)^2}}}}{\sqrt{10}-\sqrt{2}}\)

C = \(\dfrac{2\sqrt{4-\sqrt{6+\sqrt{20}}}}{\sqrt{10}-\sqrt{2}}\) = \(\dfrac{2\sqrt{4-\sqrt{\left(\sqrt{5}+1\right)^2}}}{\sqrt{10}-\sqrt{2}}\)

C = \(\dfrac{2\sqrt{3-\sqrt{5}}}{\sqrt{10}-\sqrt{2}}\) = \(\dfrac{2\sqrt{3-\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)}{10-2}\)

C = \(\dfrac{2\sqrt{30-10\sqrt{5}}+2\sqrt{6-2\sqrt{5}}}{8}\)

C = \(\dfrac{2\sqrt{\left(5-\sqrt{5}\right)^2}+2\sqrt{\left(\sqrt{5}-1\right)^2}}{8}\)

C = \(\dfrac{2\left(5-\sqrt{5}\right)+2\left(\sqrt{5}-1\right)}{8}\)

C = \(\dfrac{10-2\sqrt{5}+2\sqrt{5}-2}{8}\) = \(\dfrac{8}{8}\) = \(1\)

D = \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)

D = \(\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)^2}\)

D = \(7-3\sqrt{5}-\left(7+3\sqrt{5}\right)\) = \(7-3\sqrt{5}-7-3\sqrt{5}\)

D = \(-6\sqrt{5}\)

A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)

A = \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\) = \(\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

A = \(\sqrt{\sqrt{5}-\sqrt{5}+1}\) = \(\sqrt{1}=1\)