Bài 1: [tex]B=(2-\sqrt{3})\sqrt{26+15\sqrt{3}}-(2+\sqrt{3})\sqrt{26-15\sqrt{3}}[/tex]
[tex]B=(2-\sqrt{3})\sqrt{26+15\sqrt{3}}-(2+\sqrt{3})\sqrt{26-15\sqrt{3}}[/tex]
[tex]\Leftrightarrow B=\frac{(2-\sqrt{3})(\sqrt{2}*\sqrt{26+15\sqrt{3}})-(2+\sqrt{3})(\sqrt{2}*\sqrt{26-15\sqrt{3}})}{\sqrt{2}}[/tex]
[tex]\Leftrightarrow B=\frac{(2-\sqrt{3})\sqrt{52+30\sqrt{3}}-(2+\sqrt{3})\sqrt{52-30\sqrt{3}}}{\sqrt{2}}[/tex]
[tex]\Leftrightarrow B=\frac{(2-\sqrt{3})\sqrt{27+30\sqrt{3}+25}-(2+\sqrt{3})\sqrt{27-30\sqrt{3}+25}}{\sqrt{2}}[/tex]
[tex]\Leftrightarrow B=\frac{(2-\sqrt{3})\sqrt{(3\sqrt{3}+5)^{2}}-(2+\sqrt{3})\sqrt{(3\sqrt{3}-5)^{2}}}{\sqrt{2}}[/tex]
[tex]\Leftrightarrow B=\frac{(2-\sqrt{3})(3\sqrt{3}+5)-(2+\sqrt{3})\left |3\sqrt{3}-5 \right |}{\sqrt{2}}[/tex]
[tex]\Leftrightarrow B=\frac{(2-\sqrt{3})(3\sqrt{3}+5)-(2+\sqrt{3})(3\sqrt{3}-5)}{\sqrt{2}}[/tex] (vì [tex]3\sqrt{3}> 5[/tex] )
[tex]\Leftrightarrow B=\frac{(6\sqrt{3}+10-9-5\sqrt{3})-(6\sqrt{3}-10+9-5\sqrt{3})}{\sqrt{2}}[/tex]
[tex]\Leftrightarrow B=\frac{6\sqrt{3}+10-9-5\sqrt{3}-6\sqrt{3}+10-9+5\sqrt{3}}{\sqrt{2}}[/tex]
[tex]\Leftrightarrow B=\frac{2}{\sqrt{2}}[/tex]
[tex]\Leftrightarrow B=\sqrt{2}[/tex]
Vậy [tex]B=\sqrt{2}[/tex]