Exercitiul 6 va rog, mi se pare complicat.Multumesc anticipat​

Răspuns:

a = -6

Explicație pas cu pas:

[tex]a=\sqrt{24}\cdot \Big(\frac{3}{\sqrt{2}} - \frac{2}{\sqrt{3}} + \frac{1}{\sqrt{6}}\Big)-2(3\sqrt{3}-2\sqrt{2}+4) \\\\ = 2\sqrt{6}\cdot \frac{3\sqrt{3} - 2\sqrt{2} + 1}{\sqrt{6}} - 6\sqrt{3} + 4\sqrt{2} - 8\\\\ = 2(3\sqrt{3}-2\sqrt{2}+1)-6\sqrt{3} + 4\sqrt{2} - 8 \\\\= 6\sqrt{3} - 4\sqrt{2}+2-6\sqrt{3}+4\sqrt{2}-8 \\\\= 2-8\\\\ = -6\implies a \in \mathbb{Q}[/tex]

Răspuns:

Explicație pas cu pas:

a=[tex]a=\sqrt{24}(\frac{3}{\sqrt{2}}-\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{6} })-2(3\sqrt{3}-2\sqrt{2}+4)= \\ =\sqrt{4*6}(\frac{3}{\sqrt{2}}-\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{6} })-6\sqrt{3}+4\sqrt{2}-8=\\ =\frac{2\sqrt{6} *3}{\sqrt{2}}-\frac{2\sqrt{6} *2}{\sqrt{3}}+\frac{2\sqrt{6} *1}{\sqrt{6}}-6\sqrt{3}+4\sqrt{2}-8=6*\sqrt{\frac{6}{2}}-4*\sqrt{\frac{6}{3}}+2*\sqrt{\frac{6}{6}}-6\sqrt{3}+4\sqrt{2}-8=6\sqrt{3}-4\sqrt{2}+2-6\sqrt{3}+4\sqrt{2}-8=-6[/tex]

-6∈Q