Question

Rationalise 1/3-root7+root3

Answer 1

-\frac{1}{20}(2\sqrt{7}-6\sqrt{3}+8\sqrt{2}-2\sqrt{42})
Step-by-step explanation:
Here, the given expression is,
\frac{1}{\sqrt{7} +\sqrt{3} -\sqrt{2} }
Multiply and divide by √7 + √3 + √2,
=\frac{1}{\sqrt{7} +\sqrt{3} -\sqrt{2} }\times \frac{\sqrt{7} +\sqrt{3} +\sqrt{2}}{\sqrt{7} +\sqrt{3} +\sqrt{2}}
=\frac{\sqrt{7} +\sqrt{3} +\sqrt{2}}{(\sqrt{7} +\sqrt{3})^2-(\sqrt{2})^2 }
=\frac{\sqrt{7} +\sqrt{3} +\sqrt{2}}{7+3+2\sqrt{21}-2}
=\frac{\sqrt{7} +\sqrt{3} +\sqrt{2}}{8+2\sqrt{21}}
Multiply and divide by 8-2√21,
=\frac{\sqrt{7} +\sqrt{3} +\sqrt{2}}{8+2\sqrt{21}}\times \frac{8-2\sqrt{21}}{8-2\sqrt{21}}
=\frac{8\sqrt{7} +8\sqrt{3} +9\sqrt{2}-2\sqrt{147} -2\sqrt{63} -2\sqrt{42}}{(8)^2-(2\sqrt{21})^2}
=\frac{8\sqrt{7} +8\sqrt{3} +9\sqrt{2}-14\sqrt{3} -6\sqrt{7} -2\sqrt{42}}{64-84}
=\frac{8\sqrt{7} +8\sqrt{3} +9\sqrt{2}-14\sqrt{3} -6\sqrt{7} -2\sqrt{42}}{-20}
=\frac{2\sqrt{7} -6\sqrt{3} +9\sqrt{2}-2\sqrt{42}}{-20}
=-\frac{1}{20}(2\sqrt{7}-6\sqrt{3}+8\sqrt{2}-2\sqrt{42})

Answer 2

$\huge\underline{ \underline{ \bf{ \blue{ \: solution \: : = }}}} </p><p>$

$\frac{1}{3 - \sqrt{7} + \sqrt{3} } \\ \\ \frac{1}{3 - \sqrt{7} + \sqrt{3} } \times \frac{3 - \sqrt{7} - \sqrt{3} }{3 - \sqrt{7} - \sqrt{3} }$

solution was in the attachment.