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Fazendo √2+√3+√5 = x temos :
![\dfrac{3*(x+2)}{2*[ (x+1)^{2}-1 ]} - \dfrac{1}{x} = \dfrac{3x+6}{2*[ x^{2} +2x+1-1]} - \dfrac{1}{x} \\ \\ \\ \dfrac{3x+6}{2 x^{2} +4x} - \dfrac{1}{x} = \dfrac{3x+6-(2x+4)}{2 x^{2} +4x}= \dfrac{3x+6-2x-4}{2 x^{2} +4x}= \\ \\ \\ \dfrac{x+2}{2x(x+2)}= \dfrac{1}{2x} \quad\quad voltando\quad aos\quad radicais \\ \\ \\ \boxed { \dfrac{1}{2*( \sqrt{2}+ \sqrt{3}+ \sqrt{5} )}} }](https://tex.z-dn.net/?f=+%5Cdfrac%7B3%2A%28x%2B2%29%7D%7B2%2A%5B+%28x%2B1%29%5E%7B2%7D-1+%5D%7D+-+%5Cdfrac%7B1%7D%7Bx%7D+%3D+%5Cdfrac%7B3x%2B6%7D%7B2%2A%5B+x%5E%7B2%7D+%2B2x%2B1-1%5D%7D+-+%5Cdfrac%7B1%7D%7Bx%7D++%5C%5C++%5C%5C+%5C%5C+++%5Cdfrac%7B3x%2B6%7D%7B2+x%5E%7B2%7D+%2B4x%7D+-+%5Cdfrac%7B1%7D%7Bx%7D+%3D+%5Cdfrac%7B3x%2B6-%282x%2B4%29%7D%7B2+x%5E%7B2%7D+%2B4x%7D%3D+%5Cdfrac%7B3x%2B6-2x-4%7D%7B2+x%5E%7B2%7D+%2B4x%7D%3D+%5C%5C++%5C%5C++%5C%5C++%5Cdfrac%7Bx%2B2%7D%7B2x%28x%2B2%29%7D%3D++%5Cdfrac%7B1%7D%7B2x%7D+%5Cquad%5Cquad+voltando%5Cquad+aos%5Cquad+radicais+%5C%5C++%5C%5C++%5C%5C+%5Cboxed++%7B+%5Cdfrac%7B1%7D%7B2%2A%28+%5Csqrt%7B2%7D%2B+%5Csqrt%7B3%7D%2B+%5Csqrt%7B5%7D+++%29%7D%7D+%7D "\dfrac{3*(x+2)}{2*[ (x+1)^{2}-1 ]} - \dfrac{1}{x} = \dfrac{3x+6}{2*[ x^{2} +2x+1-1]} - \dfrac{1}{x} \\ \\ \\ \dfrac{3x+6}{2 x^{2} +4x} - \dfrac{1}{x} = \dfrac{3x+6-(2x+4)}{2 x^{2} +4x}= \dfrac{3x+6-2x-4}{2 x^{2} +4x}= \\ \\ \\ \dfrac{x+2}{2x(x+2)}= \dfrac{1}{2x} \quad\quad voltando\quad aos\quad radicais \\ \\ \\ \boxed { \dfrac{1}{2*( \sqrt{2}+ \sqrt{3}+ \sqrt{5} )}} }")
Fazendo √2+√3+√5 = x temos :
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