Question

rationlise the denominator of 1/root7+root3-root2​

Answer 1

Answer:

-\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})