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Ta có
\(\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)^2\)
\(=27+10\sqrt{2}+27-10\sqrt{2}-2\sqrt{\left(27+10\sqrt{2}\right)\left(27-10\sqrt{2}\right)}\)
\(=54-2\sqrt{529}=8\)
\(\Rightarrow\) \(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}=\sqrt{8}=2\sqrt{2}\)
Xét tử số
\(\left(27+10\sqrt{2}\right)\sqrt{27-10\sqrt{2}}-\left(27-10\sqrt{2}\right)\sqrt{27+10\sqrt{2}}\)
\(=\left(\sqrt{27+10\sqrt{2}}.\sqrt{27-10\sqrt{2}}\right)\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)\)
\(=23\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)\)
\(=23.2\sqrt{2}=46\sqrt{2}\)
Lại có \(\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right)^2\)
\(=\sqrt{13}-3+\sqrt{13}+3+2\sqrt{\left(\sqrt{13}-3\right)\left(\sqrt{13}+3\right)}\)
\(=2\sqrt{13}+2\sqrt{4}=2\sqrt{13}+4\)
ta bình phương mẫu số
\(\left(\frac{\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}}{\sqrt{\sqrt{13}+2}}\right)^2=\frac{\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right)^2}{\sqrt{13}+2}\)
\(=\frac{2\sqrt{13}+4}{\sqrt{13}+2}=2\)
Vậy mẫu \(=\sqrt{2}\)
Vậy \(x=\frac{46\sqrt{2}}{\sqrt{2}}=46\) thay vào ta đc A = 92880
\(x=\frac{\left(5+\sqrt{2}\right)^2\sqrt{\left(5-\sqrt{2}\right)^2}-\left(5-\sqrt{2}\right)^2\sqrt{\left(5+\sqrt{2}\right)^2}}{\frac{\sqrt{\left(\sqrt{13}-3\right)\left(\sqrt{13}-2\right)}+\sqrt{\left(\sqrt{13}+3\right)\left(\sqrt{13}-2\right)}}{\sqrt{13-4}}}\)
\(=\frac{\left(5+\sqrt{2}\right)\left(5+\sqrt{2}\right)\left(5-\sqrt{2}\right)-\left(5-\sqrt{2}\right)\left(5-\sqrt{2}\right)\left(5+\sqrt{2}\right)}{\frac{\sqrt{19-5\sqrt{13}}+\sqrt{7+\sqrt{13}}}{3}}\)
\(=\frac{69\left(5+\sqrt{2}-5+\sqrt{2}\right)}{\frac{1}{\sqrt{2}}\left(\sqrt{38-10\sqrt{13}}+\sqrt{14+2\sqrt{13}}\right)}=\frac{276}{\sqrt{\left(5-\sqrt{13}\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}}\)
\(=\frac{276}{5-\sqrt{13}+\sqrt{13}+1}=46\)
\(\Rightarrow A=...\)
\(\dfrac{\sqrt{3+\sqrt{5}}\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}=\dfrac{\sqrt{5+2\sqrt{5}+1}\left(\sqrt{3}+1\right)\sqrt{2}\left(\sqrt{5}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3-\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}}=\dfrac{\sqrt{2}\left(\sqrt{5}+1\right)^2\left(\sqrt{3}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3-\sqrt{3-2\sqrt{3}+1}}}=\dfrac{\left(6+2\sqrt{5}\right)\left(\sqrt{3}+1\right)\left(3-\sqrt{5}\right)}{\sqrt{2}.\sqrt{4-\sqrt{3}}}=\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{\sqrt{4-\sqrt{3}}}\)
\(=\dfrac{4\sqrt{2}\left(\sqrt{3}+1\right)}{\sqrt{4-\sqrt{3}}}\)
d/ \(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}\)
\(\Leftrightarrow x^3=6-5x\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow x=1\)
c/
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=3-1=2\)
a: \(=\sqrt{15}-3+\sqrt{15}=2\sqrt{15}-3\)
b: \(=\left(\sqrt{7}+\sqrt{3}-\sqrt{7}+\sqrt{3}\right)-2\sqrt{3}+2\)
\(=2\)
c: \(=\left(\sqrt{3}-\sqrt{2}\right)\cdot3\cdot\left(\sqrt{3}+\sqrt{2}\right)=3\)