Răspuns: [tex]\bf 4\sqrt{3} + 1[/tex]
Explicație pas cu pas:
Salutare!
[tex]\bf \sqrt{6}\cdot\Big(\dfrac{4}{\sqrt{2}}- \dfrac{3}{\sqrt{3}}\Big)+\Big( \dfrac{6}{\sqrt{12}}+\dfrac{9}{\sqrt{27}}-\dfrac{16}{4\sqrt{8}} \Big)\cdot \dfrac{1}{2\sqrt{2}-\sqrt{2}}+\sqrt{18} =[/tex]
[tex]\bf \sqrt{6}\cdot\Big(2\sqrt{2}- \sqrt{3}\Big)+\Big( \dfrac{6}{2\sqrt{3}}+\dfrac{9}{3\sqrt{3}}-\dfrac{16}{8\sqrt{2}} \Big)\cdot \dfrac{1}{2\sqrt{2}-\sqrt{2}}+3\sqrt{2} =[/tex]
[tex]\bf \Big(2\sqrt{12}- \sqrt{18}\Big)+\Big( \dfrac{3}{\sqrt{3}}+\dfrac{3}{\sqrt{3}}-\dfrac{2}{2\sqrt{2}} \Big)\cdot \dfrac{1}{2\sqrt{2}-\sqrt{2}}+3\sqrt{2} =[/tex]
[tex]\bf \Big(4\sqrt{3}- 3\sqrt{2}\Big)+\Big( \dfrac{6}{\sqrt{3}}-2\sqrt{2} \Big)\cdot \dfrac{1}{2\sqrt{2}-\sqrt{2}}+3\sqrt{2} =[/tex]
[tex]\bf \Big(4\sqrt{3}- 3\sqrt{2}\Big)+\Big( 2\sqrt{3}-2\sqrt{2} \Big)\cdot \dfrac{1}{2\sqrt{2}-\sqrt{2}}+3\sqrt{2} =[/tex]
[tex]\bf \Big(4\sqrt{3}- 3\sqrt{2}\Big)+\not\Big( 2\sqrt{3}-2\sqrt{2} \Big)\cdot \dfrac{1}{\not\Big( 2\sqrt{3}-2\sqrt{2} \Big)}+3\sqrt{2} =[/tex]
[tex]\bf 4\sqrt{3}- 3\sqrt{2}+ 1 + 3\sqrt{2} =[/tex]
[tex]\boxed{\bf 4\sqrt{3} + 1}[/tex]
#copaceibrainly
