Освободиться от иррациональности в знаменателе дроба: 3+√2+√3 / 3-√2-√3. Created by kolosil. algebra-ru

3-√2-√3...

kolosil @kolosil

June 2022 1 7 Report

Освободиться от иррациональности в знаменателе дроба: 3+√2+√3 / 3-√2-√3

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m11m

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$= \frac{(3+ \sqrt{2}+ \sqrt{3} )(3- \sqrt{2} + \sqrt{3} )}{(3- \sqrt{2} - \sqrt{3} )(3- \sqrt{2} + \sqrt{3} )} = \frac{(3+ \sqrt{3}+ \sqrt{2})(3+ \sqrt{3}- \sqrt{2} ) }{(3- \sqrt{2} )^2 -( \sqrt{3} )^2}= \\ 
 \\ 
= \frac{(3+ \sqrt{3} )^2-( \sqrt{2} )^2}{9-6 \sqrt{2}+2-3 }= \frac{9+6 \sqrt{3}+3-2 }{8-6 \sqrt{2} }= \frac{10+6 \sqrt{3} }{8-6 \sqrt{2} }= \\ 
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= \frac{2(5+3 \sqrt{3} )}{2(4-3 \sqrt{2} )}= \frac{5+3 \sqrt{3} }{4-3 \sqrt{2} }= \\ 
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= \frac{(5+3 \sqrt{3} )(4+3 \sqrt{2} )}{(4-[tex] \sqrt{2}(4+3 \sqrt{2} )}= \frac{20+12 \sqrt{3}+15 \sqrt{2}+9 \sqrt{6} }{4^2-(3 \sqrt{2} )^2}= \\ 
 \\ 
= \frac{20+12 \sqrt{3}+15 \sqrt{2}+9 \sqrt{6} }{16-18}= -10 -6 \sqrt{3} -7.5 \sqrt{2}-4.5 \sqrt{6}$ )()} [/tex]

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sravnite a2log 4 pri osnove 3 v 3log 17 pri osnove 27

razlozhit mnogochleny na mnozhiteli r3h h3 x2 8x 6

sravnite chisla a2log4 pri osnove 3 i vlog17 pri osnove 27

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