1) a) 5\(\sqrt{48}\) -4\(\sqrt{27}-2\sqrt{75}\)
=5\(\sqrt{3.6}-4\sqrt{3.9}-2\sqrt{3.25}\)
=\(\sqrt{3}.\left(5\sqrt{16}-4\sqrt{9}-2\sqrt{25}\right)\)
=\(\sqrt{3}\left(20-12-10\right)\)
=\(\sqrt{3}-2\)
b)\(\sqrt{37-20\sqrt{3}}+\sqrt{21+12\sqrt{3}}\)
=\(\sqrt{25-2.5.2\sqrt{3}+12}+\sqrt{9+2.3.2\sqrt{3}+12}=\sqrt{\left(5-2\sqrt{3}\right)^2}+\sqrt{\left(3+2\sqrt{3}\right)^2}=\left|5-2\sqrt{3}\right|+\left|3+2\sqrt{3}\right|=5-2\sqrt{3}+3+2\sqrt{3}=8\)
c)=\(\left|\sqrt{13}-5\right|+\left|\sqrt{13}+7\right|=5-\sqrt{13}+\sqrt{13}+7=12\)
d)=\(\dfrac{3\left(\sqrt{2}+1\right)+3\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
=\(\dfrac{3\sqrt{2}}{\left(\sqrt{2}\right)^2-1}\)
=\(3\sqrt{2}\)
3)<=>\(\sqrt{\left(x-3\right)^2}=6\)
<=>\(\left|x-3\right|=6\)
<=>\(\left[{}\begin{matrix}x-3=6\Leftrightarrow x=9\\x-3=-6\leftrightarrow x=-3\end{matrix}\right.\)
vậy pt có 2 nghiệm x=9;x=-3
