Question

2√2supra 3√3 + 5√2supra2√3- (7√2supra6√3+ 4√3supra√3)
Ajutatima va rog

Answer 1

Aduci la acelasi numitor.
$\frac{2 \sqrt{2} }{3 \sqrt{3} } + \frac{5 \sqrt{2} }{2 \sqrt{3} } - ( \frac{7 \sqrt{2} }{6 \sqrt{3} } + \frac{4 \sqrt{3} }{\sqrt{3} }) = \frac{4 \sqrt{2} }{6 \sqrt{3} } + \frac{15 \sqrt{2} }{6 \sqrt{3} } - ( \frac{7 \sqrt{2} }{6 \sqrt{3} } + \frac{24 \sqrt{3} }{6 \sqrt{3} }) = \frac{19 \sqrt{2} }{6 \sqrt{3} } - \frac{7 \sqrt{2} + 24 \sqrt{3} }{6 \sqrt{3} } = \frac{19 \sqrt{2} - 7 \sqrt{2} - 24 \sqrt{3}}{6 \sqrt{3} } = \frac{12 \sqrt{2} - 24 \sqrt{3}}{6 \sqrt{3} } = \frac{6(2 \sqrt{2} - 4 \sqrt{3})}{6 \sqrt{3} } =$$\frac{2 \sqrt{2} - 4 \sqrt{3}}{ \sqrt{3} }$
Rationalizezi si o sa obtii:
$\frac{2 \sqrt{2} - 4 \sqrt{3}}{ \sqrt{3} } = \frac{2 \sqrt{6} - 12}{3}$