\(\sqrt{\sqrt{3}+2+\sqrt{7-4\sqrt{3}}}=\sqrt{\sqrt{3}+2+2-\sqrt{3}}=\sqrt{4}=2\)LÀ MỘT SỐ NGUYÊN

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

<=>4 là số nguyên => t là số nguyên

\(A=\left(\dfrac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}-\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right)\left(\dfrac{\sqrt{3}-1}{3\sqrt{2}-\sqrt{6}}\right)\)

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

=-4

=5+2√6−5+2√6−1⋅1√6=5+26−5+26−1⋅16

=-4

\(BT=\sqrt{6+2\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}\)

\(=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}\)

\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)

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

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

\(=1+\sqrt{3}\)

Ta có: \(a+b\sqrt{3}=\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)

\(\Leftrightarrow a+b\sqrt{3}=2-\sqrt{3}-2-\sqrt{3}\)

\(\Leftrightarrow a+b\sqrt{3}=-2\sqrt{3}\)

\(\Leftrightarrow a=0;b=-2\)

T=a+b=0+(-2)=-2

\(S=\sqrt{\left(\sqrt{3}\right)^2-2\cdot2\sqrt{3}+2^2}-\sqrt{\left(\sqrt{3}\right)^2+2\cdot2\cdot\sqrt{3}+2^2}\)

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

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

\(a=0,b=-2\)

\(T=0+-2=-2\)

\(\sqrt{3}+2+\sqrt{7-4\sqrt{3}}=\sqrt{3}+2+\sqrt{4-2.2\sqrt{3}+3}\)

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

=>ĐPCM