Question

simplify by rationalising the denominator: 1 / root7+root6-root3​

Answer 1

Given,

$\longrightarrow x=\dfrac{1}{\sqrt7+\sqrt6-\sqrt3}$

We've to rewrite $x$ in simplified form by rationalising it.

Multiplying both numerator and denominator by $\sqrt7+\sqrt6+\sqrt3,$

$\longrightarrow x=\dfrac{\sqrt7+\sqrt6+\sqrt3}{(\sqrt7+\sqrt6-\sqrt3)(\sqrt7+\sqrt6+\sqrt3)}$

$\longrightarrow x=\dfrac{\sqrt7+\sqrt6+\sqrt3}{(\sqrt7+\sqrt6)^2-(\sqrt3)^2}$

$\longrightarrow x=\dfrac{\sqrt7+\sqrt6+\sqrt3}{7+6+2\sqrt{42}-3}$

$\longrightarrow x=\dfrac{\sqrt7+\sqrt6+\sqrt3}{10+2\sqrt{42}}$

$\longrightarrow x=\dfrac{\sqrt7+\sqrt6+\sqrt3}{2(5+\sqrt{42})}$

$\longrightarrow 2x=\dfrac{\sqrt7+\sqrt6+\sqrt3}{5+\sqrt{42}}$

Now, multiplying both numerator and denominator by $5-\sqrt{42},$

$\longrightarrow 2x=\dfrac{(\sqrt7+\sqrt6+\sqrt3)(5-\sqrt{42})}{(5+\sqrt{42})(5-\sqrt{42})}$

$\longrightarrow 2x=\dfrac{5(\sqrt7+\sqrt6+\sqrt3)-\sqrt{42}(\sqrt7+\sqrt6+\sqrt3)}{5^2-(\sqrt{42})^2}$

$\longrightarrow 2x=\dfrac{5\sqrt7+5\sqrt6+5\sqrt3-7\sqrt6-6\sqrt7-3\sqrt{14}}{25-42}$

$\longrightarrow 2x=\dfrac{-\sqrt7-2\sqrt6+5\sqrt3-3\sqrt{14}}{-17}$

$\longrightarrow x=\dfrac{\sqrt7+2\sqrt6+3\sqrt{14}-5\sqrt3}{34}$

$\longrightarrow\underline{\underline{x=\dfrac{(1+3\sqrt2)\sqrt7+(2\sqrt2-5)\sqrt3}{34}}}$