Question

Va rog ex28si29.Dau coroana!!!

Answer 1

$\displaystyle \mathtt{28a) \frac{3}{5+2 \sqrt{6} }- \frac{4}{5 \sqrt{2}-4 \sqrt{3}}+ \frac{18}{3- \sqrt{6} }+ \frac{24}{3 \sqrt{2}-2 \sqrt{3}}=}\\ \\ \mathtt{= \frac{3 \left(5-2 \sqrt{6}\right) }{25-24} - \frac{4\left(5 \sqrt{2}+4 \sqrt{3}\right)}{50-48} + \frac{18\left(3+ \sqrt{6}\right) }{9-6} + \frac{24 \left(3 \sqrt{2}+2 \sqrt{3}\right)}{18-12}= }$
$\displaystyle \mathtt{=15-6 \sqrt{6} - \frac{20 \sqrt{2}+16 \sqrt{3} }{2}+ \frac{54+18 \sqrt{6} }{3}+ \frac{72 \sqrt{2}+ 48 \sqrt{3} }{6}= }\\ \\ \mathtt{= \frac{90-36 \sqrt{6}-3 \left(20 \sqrt{2}+16 \sqrt{3} \right)+2\left(54+18 \sqrt{6}\right)+72 \sqrt{2}+48 \sqrt{3}}{6} =} \\ \\ \mathtt{= \frac{90-36 \sqrt{6}-60 \sqrt{2}- 48 \sqrt{3}+108+36 \sqrt{6} +72 \sqrt{2}+48 \sqrt{3}}{6}=}\\ \\ \mathtt{= \frac{198+12 \sqrt{2} }{6} = \frac{\not6 \left(33+2 \sqrt{2}\right) }{\not 6} =33+2 \sqrt{2} }$
$\displaystyle \mathtt{b) \frac{2}{3+2 \sqrt{2} }+ \frac{2}{3 \sqrt{2} -4} + \frac{4}{6+4 \sqrt{2}}+ \frac{4}{5 \sqrt{2}+7 } =}\\ \\ \mathtt{= \frac{2\left(3-2 \sqrt{2}\right)}{9-8}+ \frac{2 \left(3 \sqrt{2}+4\right) }{18-16}+ \frac{4 \left(6-4 \sqrt{2}\right) }{36-32}+ \frac{4\left(5 \sqrt{2} -7 \right)}{50-49}=}$
$\displaystyle \mathtt{=6-4 \sqrt{2}+ \frac{\not2\left(3 \sqrt{2}+4 \right) }{\not 2}+ \frac{\not4\left(6-4 \sqrt{2}\right) }{\not4} +20 \sqrt{2}-28= }\\ \\ \mathtt{=6-4 \sqrt{2}+3 \sqrt{2}+4+6-4 \sqrt{2}+20 \sqrt{2}-28=15 \sqrt{2} -12 }$
$\displaystyle \mathtt{c) \frac{6}{2 \sqrt{7}+5}- \frac{4}{3- \sqrt{7} } + \frac{2}{8+3 \sqrt{7}}+ \frac{9}{11-4 \sqrt{7}}=}\\ \\ \mathtt{ =\frac{6 \left(2 \sqrt{7}-5\right) }{28-25} - \frac{4 \left(3+ \sqrt{7}\right) }{9-7} + \frac{2 \left(8-3 \sqrt{7}\right) }{64-63}+ \frac{9 \left(11+4 \sqrt{7}\right) }{121-112}=}$
[tex] \displaystyle \mathtt{= \frac{\not6\left(2 \sqrt{7}-5\right) }{\not3} - \frac{\not4\left(3+ \sqrt{7}\right) }{\not 2}+16-6 \sqrt{7} + \frac{\not9 \left(11+4 \sqrt{7}\right) }{\not9}= } \\ \\ \mathtt{=2 \left(2 \sqrt{7}-5\right) -2\left(3+ \sqrt{7}\right)+16-6 \sqrt{7}+11+4 \sqrt{7} = }\\ \\ \mathtt{=4 \sqrt{7}-10-6-2 \sqrt{7}+16- 6 \sqrt{7}+11+4 \sqrt{7}= 11}[/tex]
$\displaystyle \mathtt{d) \frac{2}{ \sqrt{5}+ \sqrt{3}}+ \frac{7}{2 \sqrt{3} - \sqrt{5}}- \frac{8}{2 \sqrt{5} -2 \sqrt{3}}+ \frac{3}{4 \sqrt{3} +3 \sqrt{5} } =}$
$\displaystyle \mathtt{= \frac{2\left( \sqrt{5}- \sqrt{3}\right)}{5-3} + \frac{7 \left(2 \sqrt{3}+ \sqrt{5}\right)}{12-5} - \frac{8 \left(2 \sqrt{5}+2 \sqrt{3}\right)}{20-12}+ \frac{3 \left(4 \sqrt{3}-3 \sqrt{5}\right)}{48-45}=}$
$\displaystyle \mathtt{ \frac{\not2\left( \sqrt{5}- \sqrt{3}\right) }{\not2} + \frac{\not7\left(2 \sqrt{3}+ \sqrt{5}\right) }{\not7} - \frac{\not8 \left(2 \sqrt{5}+2 \sqrt{3}\right) }{\not8}+ \frac{\not3\left(4 \sqrt{3}-3 \sqrt{5}\right)}{\not3} }\\ \\ \mathtt{= \sqrt{5}- \sqrt{3}+2 \sqrt{3}+ \sqrt{5}-2 \sqrt{5}-2 \sqrt{3}+4 \sqrt{3}-3 \sqrt{5}=3 \sqrt{3}-3 \sqrt{5}}$