Bài 1:
a: \(2\cdot\sqrt{\left(1+\sqrt3\right)^2}-\sqrt{\left(2\sqrt3-3\right)^2}\)
\(=2\left(\sqrt3+1\right)-\left(2\sqrt3-3\right)\)
\(=2\sqrt3+2-2\sqrt3+3=5\)
b: \(\left(1+\sqrt3-\sqrt5\right)\left(1+\sqrt3+\sqrt5\right)\)
\(=\left(1+\sqrt3\right)^2-5\)
\(=4+2\sqrt3-5=2\sqrt3-1\)
c: \(\left(\sqrt{\frac83}-\sqrt{24}+\sqrt{\frac{50}{3}}\right)\cdot\sqrt6\)
\(=\left(\frac{\sqrt{24}}{3}-2\sqrt6+\frac{\sqrt{150}}{3}\right)\cdot\sqrt6=\left(\frac{2\sqrt6}{3}-2\sqrt6+\frac{5\sqrt6}{3}\right)\cdot\sqrt6\)
\(=\frac13\sqrt6\cdot\sqrt6=\frac13\cdot6=2\)
e: \(\sqrt{3+2\sqrt2}-\sqrt{3-2\sqrt2}\)
\(=\sqrt{\left(\sqrt2+1\right)^2}-\sqrt{\left(\sqrt2-1\right)^2}\)
\(=\sqrt2+1-\left(\sqrt2-1\right)=2\)
Bài 2:
a: \(\sqrt{64a^2}+2a=8\cdot\left|a\right|+2a=8a+2a\left(a\ge0\right)\)
=10a
b: \(3\cdot\sqrt{9a^{10}}+6a^5\)
\(=3\cdot3\cdot\left|a^5\right|+6a^5=9\cdot\left|a^5\right|+6a^5\)
\(=-9a^5+6a^5=-3a^5\)
c: \(\sqrt{50\cdot18\left(1-x\right)^2}=\sqrt{900}\cdot\left|x-1\right|\)
=30(x-1)(x>=1)
=30x-30
