Question

освободиться от иррациональности в дроби 35\√37+√2

Answer 1

$\frac{35}{ \sqrt{37}+ \sqrt{2} } = \frac{35*( \sqrt{37}- \sqrt{2}) }{( \sqrt{37}+ \sqrt{2})( \sqrt{37} - \sqrt{2} )} = \frac{35*( \sqrt{37}- \sqrt{2}) }{( \sqrt{37}) ^{2} -( \sqrt{2}) ^{2} } = \frac{35( \sqrt{37} - \sqrt{2}) }{37-2} = \frac{35( \sqrt{37}- \sqrt{2} ) }{35}$$= \sqrt{37}- \sqrt{2}$