Efectuăm înmulțirile, simplificăm, apoi scoatem factorii de sub radicali
[tex]\sqrt{6} \bigg(3 \sqrt{3} - 4 \sqrt{2} \bigg) + \sqrt{24} \bigg( \dfrac{5}{ \sqrt{2} } - \dfrac{7}{ \sqrt{3} } \bigg) + \sqrt{50} = \\[/tex]
[tex]= \sqrt{6} \cdot 3\sqrt{3} - \sqrt{6} \cdot 4\sqrt{2} + \dfrac{5\sqrt{24}}{ \sqrt{2} } - \dfrac{7\sqrt{24}}{ \sqrt{3} } + \sqrt{2 \cdot 5^2}\\[/tex]
[tex]= 3 \sqrt{18} - 4 \sqrt{12} + 5\sqrt{12} - 7\sqrt{8} + 5\sqrt{2}\\[/tex]
[tex]= 3 \sqrt{2 \cdot 3^2} - 4 \sqrt{2^2 \cdot 3} + 5\sqrt{2^2 \cdot 3} - 7 \sqrt{2^3} + 5\sqrt{2}[/tex]
[tex]= 3 \cdot 3\sqrt{2} - 4 \cdot 2\sqrt{3} + 5 \cdot 2\sqrt{3} - 7 \cdot 2\sqrt{2} + 5\sqrt{2}[/tex]
[tex]= 9\sqrt{2} - 8\sqrt{3} + 10\sqrt{3} - 14\sqrt{2} + 5\sqrt{2}[/tex]
[tex]= (9-14+5)\sqrt{2} + (10-8)\sqrt{3}[/tex]
[tex]=\bf 2\sqrt{3}[/tex]
R: d) 2√3
