Question

[Taev Kim]

\(\sqrt{3}×\sqrt{27}-\sqrt{144}:\sqrt{36}\)

\(\left(2\sqrt{9}+3\sqrt{36}\right):4\)

\(\sqrt{7}-\sqrt{8-2\sqrt{7}}\)

\(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)

\(\dfrac{5+3\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\)

\(\sqrt{27}+5\sqrt{12}-2\sqrt{3}=11\sqrt{3}\)

Answer 1

[nguyen thi minh ngoc]

\(\sqrt{3\cdot27}-\sqrt{\dfrac{144}{36}}\)=\(\sqrt{81}-\sqrt{4}\)=9-2=7

\(\dfrac{2\cdot3+3\cdot6}{4}\)=6

\(\sqrt{7}-\sqrt{7-2\cdot\sqrt{7}+1}\)=\(\sqrt{7}-\left(\sqrt{7}-1\right)\)=1

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

\(\dfrac{\sqrt{5}\cdot\left(\sqrt{5}+3\right)}{\sqrt{5}}\)+\(\dfrac{\sqrt{3}\cdot\left(1+\sqrt{3}\right)}{\sqrt{3}+1}\)-(\(\sqrt{5}+3\))

=(\(\sqrt{5}+3\))+\(\sqrt{3}\)-(\(\sqrt{5}+3\))=\(\sqrt{3}\)

\(\sqrt{3}\cdot\sqrt{9}+5\cdot\sqrt{4}\cdot3-2\sqrt{3}\)

=\(\sqrt{3}\cdot\left(3+10-2\right)\)

=\(11\sqrt{3}\)