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Sửa đề

\(A=\left(2-\sqrt{3}\right)\sqrt[3]{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{26-15\sqrt{3}}\)

\(=\left(2-\sqrt{3}\right)\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{8-12\sqrt{3}+18-3\sqrt{3}}\)

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

\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=0\)

a)

\(A=\sqrt{26+15\sqrt{3}}=\sqrt{\frac{52+30\sqrt{3}}{2}}=\sqrt{\frac{27+25+2\sqrt{27.25}}{2}}\)

\(=\sqrt{\frac{(\sqrt{27}+\sqrt{25})^2}{2}}=\frac{\sqrt{27}+\sqrt{25}}{\sqrt{2}}=\frac{3\sqrt{3}+5}{\sqrt{2}}=\frac{3\sqrt{6}+5\sqrt{2}}{2}\)

b)

\(B\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)

\(=\sqrt{7+1+2\sqrt{7}}-\sqrt{7+1-2\sqrt{7}}-2\)

\(=\sqrt{(\sqrt{7}+1)^2}-\sqrt{(\sqrt{7}-1)^2}-2=\sqrt{7}+1-(\sqrt{7}-1)-2=0\)

\(\Rightarrow B=0\)

c)

\(C=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{3+5-2\sqrt{3.5}}-\sqrt{3+5+2\sqrt{3.5}}\)

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

d)

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

\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(2+3+2\sqrt{2.3})\sqrt{2+3-2\sqrt{2.3}}\)

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

\(=\sqrt{2}(\sqrt{3}-\sqrt{2})^2(\sqrt{3}+\sqrt{2})^2=\sqrt{2}[(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})]^2\)

\(=\sqrt{2}.1^2=\sqrt{2}\)

e)

\(E=(\sqrt{10}-\sqrt{2})\sqrt{3+\sqrt{5}}=(\sqrt{5}-1).\sqrt{2}.\sqrt{3+\sqrt{5}}\)

\(=(\sqrt{5}-1)\sqrt{6+2\sqrt{5}}=(\sqrt{5}-1)\sqrt{5+1+2\sqrt{5.1}}\)

\(=(\sqrt{5}-1)\sqrt{(\sqrt{5}+1)^2}=(\sqrt{5}-1)(\sqrt{5}+1)=4\)

f)

\(F=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-2\sqrt{20.9}}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-3)^2}}}=\sqrt{\sqrt{5}-\sqrt{3-(\sqrt{20}-3)}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}}=\sqrt{\sqrt{5}-(\sqrt{5}-1)}=\sqrt{1}=1\)

g) Áp dụng kết quả phần a):

\(G=(2-\sqrt{3}).\frac{3\sqrt{6}+5\sqrt{2}}{2}-(2+\sqrt{3}).\frac{3\sqrt{6}-5\sqrt{2}}{2}\)

\(=\frac{\sqrt{6}+\sqrt{2}}{2}-\frac{\sqrt{6}-\sqrt{2}}{2}=\sqrt{2}\)

h)

\(H=\frac{(2+\sqrt{3})\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}=\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}\)

\(=\sqrt{\frac{(2-\sqrt{3})^2}{(2+\sqrt{3})(2-\sqrt{3})}}=\sqrt{\frac{(2-\sqrt{3})^2}{2^2-3}}=\sqrt{(2-\sqrt{3})^2}=2-\sqrt{3}\)

a) Ta có: \(M=\dfrac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}\)

\(=\dfrac{2\left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}-2\sqrt{7}+3\sqrt{6}\)

\(=2\sqrt{7}+2\sqrt{6}-2\sqrt{7}+3\sqrt{6}\)

\(=5\sqrt{6}\)

b) Ta có: \(N=\left(2-\sqrt{3}\right)\left(\sqrt{26+15\sqrt{3}}\right)-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)

\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{27+2\cdot3\sqrt{3}\cdot5+25}-\left(2+\sqrt{3}\right)\sqrt{27-2\cdot3\sqrt{3}\cdot5+25}}{\sqrt{2}}\)

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

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

\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-\left(6\sqrt{3}-10+9-5\sqrt{3}\right)}{\sqrt{2}}\)

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

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

\(C=\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}=\dfrac{\left(2-\sqrt{3}\right)\sqrt{27+2.3\sqrt{3}.5+25}-\left(2+\sqrt{3}\right)\sqrt{27-2.3\sqrt{3}.5+25}}{\sqrt{2}}=\dfrac{\left(2-\sqrt{3}\right)\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-6\sqrt{3}+10-9+5\sqrt{3}}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

Ta có: \(\hept{\begin{cases}\left(2-\sqrt{3}\right)^2.\left(26+15\sqrt{3}\right)=2+\sqrt{3}\\\left(2+\sqrt{3}\right)^2.\left(26-15\sqrt{3}\right)=2-\sqrt{3}\end{cases}}\)

Sửa đề:

\(\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)

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

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

\(=\sqrt{2}\)

Xét: \(A=\sqrt{26+15\sqrt{3}}\)  dễ thấy A > 0

\(\Leftrightarrow A^2=52-2\sqrt{26^2-15^2.3}=50\Leftrightarrow A=\sqrt{50}\)

Vậy: \(A=2+\sqrt{3}.\sqrt{26+15\sqrt{3}}-2\sqrt{3}.\sqrt{26-15\sqrt{3}}\)

\(=2+\sqrt{3}.A=2+\sqrt{3}.\sqrt{50}=5\sqrt{6}+10\sqrt{2}\)

\(B^2=\left(2-\sqrt{3}\right)^2.\left(26+15\sqrt{3}\right)+\left(2+\sqrt{3}\right)^2.\left(26-15\sqrt{3}\right)-2\left(4-3\right)\sqrt{26^2-3.15^2}\)

\(B^2=\left(7-4\sqrt{3}\right).\left(26+15\sqrt{3}\right)+\left(7+4\sqrt{3}\right)\left(26-15\sqrt{3}\right)-2\)

\(B^2+2=\left(a-b\right)\left(c+d\right)+\left(a+b\right)\left(c-d\right)=ac+ad-bc-bd+ac-ad+bc-bd=2\left(ac-bd\right)\)\(B^2+2=2.\left(7.26-4.3.15\right)=2\left(182-180\right)\Rightarrow B^2=2\)

\(B>0\Rightarrow B=\sqrt{2}\)