Bài 1:
a: \(\sqrt{\left(2\sqrt2-1\right)^2}-\sqrt{17+12\sqrt2}\)
\(=\left(2\sqrt2-1\right)-\sqrt{\left(3+2\sqrt2\right)^2}\)
\(=2\sqrt2-1-3-2\sqrt2=-4\)
b: \(\left(\sqrt{10}-\sqrt2\right)\cdot\sqrt{3+\sqrt5}\)
\(=\left(\sqrt5-1\right)\cdot\sqrt{6+2\sqrt5}\)
\(=\left(\sqrt5-1\right)\cdot\sqrt{\left(\sqrt5+1\right)^2}=\left(\sqrt5-1\right)\left(\sqrt5+1\right)=5-1=4\)
c: \(\frac{2\sqrt3+3\sqrt2}{\sqrt2+\sqrt3}-4\cdot\sqrt{\frac32}-\frac{5}{1-\sqrt6}\)
\(=\frac{\sqrt6\left(\sqrt3+\sqrt2\right)}{\sqrt3+\sqrt2}-2\sqrt6+\frac{5}{\sqrt6-1}\)
\(=\sqrt6-2\sqrt6+\frac{5\left(\sqrt6+1\right)}{\left(\sqrt6-1\right)\cdot\left(\sqrt6+1\right)}=-\sqrt6+\sqrt6+1=1\)
Bài 2:
a: \(\frac{6-\sqrt6}{\sqrt6-1}-9\cdot\sqrt{\frac23}-\frac{4}{2-\sqrt6}\)
\(=\frac{\sqrt6\left(\sqrt6-1\right)}{\sqrt6-1}-3\sqrt3\cdot\sqrt2+\frac{4}{\sqrt6-2}\)
\(=\sqrt6-3\sqrt6+\frac{4\left(\sqrt6+2\right)}{\left(\sqrt6-2\right)\left(\sqrt6+2\right)}\)
\(=-2\sqrt6+2\left(\sqrt6+2\right)=-2\sqrt6+2\sqrt6+4=4\)
b: \(\left(3\sqrt2+\sqrt{10}\right)\cdot\sqrt{28-12\sqrt5}\)
\(=\sqrt2\left(3+\sqrt5\right)\cdot\sqrt{28-2\cdot\sqrt{180}}\)
\(=\sqrt2\left(3+\sqrt5\right)\cdot\sqrt{\left(3\sqrt2-\sqrt{10}\right)^2}=\sqrt2\left(3+\sqrt5\right)\left(3\sqrt2-\sqrt{10}\right)\)
\(=2\left(3+\sqrt5\right)\left(3-\sqrt5\right)=4\cdot2=8\)
