Question

8/ b…………………………………………..

Answer 1

Răspuns:

$= ( \frac{3}{ \sqrt{2} } - \frac{5 \sqrt{2} }{4} ) - ( \frac{18 \sqrt{2} }{6} + \frac{12}{3 \sqrt{2} } ) = \\ = ( \frac{3 \sqrt{2} }{2} - \frac{5 \sqrt{2} }{4} ) - ( \frac{18 \sqrt{2} }{6} + \frac{12 \sqrt{2} }{6} ) = \\ = ( \frac{6 \sqrt{2} }{4} - \frac{5 \sqrt{2} }{4} ) - \frac{30 \sqrt{2} }{6} = \\ = \frac{ \sqrt{2} }{4} - \frac{30 \sqrt{2} }{6} = \frac{6 \sqrt{2} }{24} - \frac{120 \sqrt{2} }{24} = - \frac{114 \sqrt{2} }{24} = - \frac{57 \sqrt{2} }{12} = - \frac{19 \sqrt{2} }{4}$

Answer 2

 

$\displaystyle\bf\\8b)\\\\\left(\frac{3}{\sqrt{2}}-\frac{5\sqrt{2}}{4}\right)-\left(\frac{18\sqrt{2}}{6}+\frac{12}{3\sqrt{2}}\right)=\\\\\\=\left(\frac{3\sqrt{2}}{2}-\frac{5\sqrt{2}}{4}\right)-\left(\frac{18\sqrt{2}}{6}+\frac{12\sqrt{2}}{3\times2}\right)=\\\\\\=\left(\frac{6\sqrt{2}}{4}-\frac{5\sqrt{2}}{4}\right)-\left(\frac{18\sqrt{2}}{6}+\frac{12\sqrt{2}}{6}\right)=$

.

$\displaystyle\bf\\=\frac{6\sqrt{2}-5\sqrt{2}}{4}-\left(3\sqrt{2}+2\sqrt{2}\right)=\\\\=\frac{\sqrt{2}}{4}-5\sqrt{2}=\\\\=\frac{\sqrt{2}-4\times5\sqrt{2}}{4}=\\\\=\frac{\sqrt{2}-20\sqrt{2}}{4}=\\\\=\boxed{\bf-\frac{19\sqrt{2}}{4}}$