Expressions with rationalization

Text

Evaluate, rationalizing where needed (they all give the same simple result):

• (a) $-20\sqrt{2}+\left(\dfrac{1+\sqrt{2}}{1-\sqrt{2}}\right)^2-\dfrac{16}{2+\sqrt{2}}$
• (b) $-\dfrac{1}{2}\sqrt{3}\left(\dfrac{1+\sqrt{2}}{\sqrt{3}(1-\sqrt{2})}+\dfrac{1+2\sqrt{2}}{\sqrt{3}}\right)$
• (c) $\dfrac{4}{2+\sqrt{3}}+\dfrac{\left(\sqrt{2}-\sqrt{3}\right)^2\left(\sqrt{2}+\sqrt{3}\right)^2}{\left(1-\sqrt{2}-\sqrt{6}\right)\left(1+\sqrt{2}+\sqrt{6}\right)}$

Solution

(a) Rationalize with the conjugate $1+\sqrt2$: $\dfrac{1+\sqrt2}{1-\sqrt2}=\dfrac{(1+\sqrt2)^2}{(1-\sqrt2)(1+\sqrt2)}=\dfrac{3+2\sqrt2}{-1}=-3-2\sqrt2$, so squared it is $17+12\sqrt2$. Also $\dfrac{16}{2+\sqrt2}=\dfrac{16(2-\sqrt2)}{(2+\sqrt2)(2-\sqrt2)}=\dfrac{16(2-\sqrt2)}{2}=16-8\sqrt2$. Total: $-20\sqrt2+(17+12\sqrt2)-(16-8\sqrt2)=1+(-20+12+8)\sqrt2=1$.

(b) As above $\dfrac{1+\sqrt2}{1-\sqrt2}=-3-2\sqrt2$, so the bracket is $\dfrac{-3-2\sqrt2}{\sqrt3}+\dfrac{1+2\sqrt2}{\sqrt3}=\dfrac{-2}{\sqrt3}$. Then $-\dfrac12\sqrt3\cdot\dfrac{-2}{\sqrt3}=1$.

(c) $\dfrac{4}{2+\sqrt3}=\dfrac{4(2-\sqrt3)}{4-3}=8-4\sqrt3$. Numerator: $\big[(\sqrt2-\sqrt3)(\sqrt2+\sqrt3)\big]^2=(2-3)^2=1$. Denominator: $1-(\sqrt2+\sqrt6)^2=1-(8+4\sqrt3)=-7-4\sqrt3$, so $\dfrac{1}{-7-4\sqrt3}=\dfrac{-7+4\sqrt3}{49-48}=-7+4\sqrt3$. Total: $(8-4\sqrt3)+(-7+4\sqrt3)=1$.

$\boxed{\;(a)=(b)=(c)=1\;}$