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Answered

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rationalize the denominator of √3+√2\ 5+√2 ​

Sagot :

sreedevi102

Answer:

[tex]\frac{ 5 \sqrt3 \ + \ 5 \sqrt2 \ - \ \sqrt6 \ - \ 2}{23}[/tex]

Step-by-step explanation:

[tex]\frac{\sqrt3 \ + \ \sqrt2 }{5 \ + \ \sqrt2 } \\\\=\frac{\sqrt3 \ + \ \sqrt2 }{5 \ + \ \sqrt2 } \times \frac{5 \ - \ \sqrt2 }{5 \ - \ \sqrt2 } \\\\=\frac{( \sqrt3 \ + \ \sqrt2)(5 \ - \ \sqrt2)}{(5 \ + \ \sqrt2)( 5 \ - \ \sqrt 2 )}\\\\=\frac{( \sqrt3 \ + \ \sqrt2)(5 \ - \ \sqrt2)}{(5 \ + \ \sqrt2)( 5 \ - \ \sqrt 2 )}\\\\=\frac{5 \sqrt3 \ + \ 5\sqrt 2 \ - \ \sqrt{ 3\times 2 } \ - \ \sqrt{2 \times 2}}{(5)^2 \ - \ (\sqrt2)^2}\\\\= \frac{ 5 \sqrt3 \ + \ 5 \sqrt2 \ - \ \sqrt6 \ - \ 2}{25 - 2}\\\\[/tex]

[tex]= \frac{ 5 \sqrt3 \ + \ 5 \sqrt2 \ - \ \sqrt6 \ - \ 2}{23}[/tex]

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