Bài 1:
a: ĐKXĐ: 3x-2>=0
=>3x>=2
=>\(x\ge\frac23\)
b: TA có: \(\sqrt{50}-\sqrt8+\sqrt{18}-4\cdot\sqrt{32}\)
\(=5\sqrt2-2\sqrt2+3\sqrt2-4\cdot4\sqrt2\)
\(=6\sqrt2-16\sqrt2=-10\sqrt2\)
Bài 2:
a: \(\left(3-\sqrt2\right)\cdot\sqrt{11+6\sqrt2}\)
\(=\left(3-\sqrt2\right)\cdot\sqrt{\left(3+\sqrt2\right)^2}\)
\(=\left(3-\sqrt2\right)\left(3+\sqrt2\right)=9-2=7\)
b: \(\sqrt2\cdot\sqrt{7+3\sqrt5}-\frac{4}{\sqrt5-1}\)
\(=\sqrt{14+6\sqrt5}-\frac{4\left(\sqrt5+1\right)}{\left(\sqrt5-1\right)\left(\sqrt5+1\right)}\)
\(=\sqrt{\left(3+\sqrt5\right)^2}-\left(\sqrt5+1\right)=3+\sqrt5-\sqrt5-1=3-1=2\)
c: \(\sqrt{27}-6\cdot\sqrt{\frac13}-\frac{\sqrt3-3}{\sqrt3}\)
\(=3\sqrt3-2\sqrt3-\left(1-\sqrt3\right)=\sqrt3+\sqrt3-1=2\sqrt3-1\)
d: \(\frac{9-2\sqrt3}{3\sqrt6-2\sqrt2}=\frac{\sqrt3\left(3\sqrt3-2\right)}{\sqrt2\left(3\sqrt3-2\right)}=\sqrt{\frac32}=\frac{\sqrt6}{2}\)
