7327{\displaystyle {\frac {7{\sqrt {3}}}{2{\sqrt {7}}}}} We can see that there is a 7{\displaystyle {\sqrt {7}}} in the denominator.
7327⋅77{\displaystyle {\frac {7{\sqrt {3}}}{2{\sqrt {7}}}}\cdot {\frac {\sqrt {7}}{\sqrt {7}}}}
7327⋅77=72114=212{\displaystyle {\frac {7{\sqrt {3}}}{2{\sqrt {7}}}}\cdot {\frac {\sqrt {7}}{\sqrt {7}}}={\frac {7{\sqrt {21}}}{14}}={\frac {\sqrt {21}}{2}}}
42+2{\displaystyle {\frac {4}{2+{\sqrt {2}}}}} To see why this is the case, write an arbitrary fraction 1a+b,{\displaystyle {\frac {1}{a+b}},} where a{\displaystyle a} and b{\displaystyle b} are irrational. Then the expression (a+b)(a+b)=a2+2ab+b2{\displaystyle (a+b)(a+b)=a^{2}+2ab+b^{2}} contains a cross-term 2ab. {\displaystyle 2ab. } If at least one of a{\displaystyle a} and b{\displaystyle b} is irrational, then the cross-term will contain a radical. Let’s see how this works with our example. 42+2⋅2+22+2=4(2+2)4+42+2{\displaystyle {\frac {4}{2+{\sqrt {2}}}}\cdot {\frac {2+{\sqrt {2}}}{2+{\sqrt {2}}}}={\frac {4(2+{\sqrt {2}})}{4+4{\sqrt {2}}+2}}} As you can see, there’s no way we can get rid of the 42{\displaystyle 4{\sqrt {2}}} in the denominator after doing this.
42+2⋅2−22−2{\displaystyle {\frac {4}{2+{\sqrt {2}}}}\cdot {\frac {2-{\sqrt {2}}}{2-{\sqrt {2}}}}} Why does the conjugate work? Going back to our arbitrary fraction 1a+b,{\displaystyle {\frac {1}{a+b}},} multiplying by the conjugate in the numerator and denominator results in the denominator being (a+b)(a−b)=a2−b2. {\displaystyle (a+b)(a-b)=a^{2}-b^{2}. } The key here is that there are no cross-terms. Since both of these terms are being squared, any square roots will be eliminated.
42+2⋅2−22−2=4(2−2)4−2=4−22{\displaystyle {\frac {4}{2+{\sqrt {2}}}}\cdot {\frac {2-{\sqrt {2}}}{2-{\sqrt {2}}}}={\frac {4(2-{\sqrt {2}})}{4-2}}=4-2{\sqrt {2}}}
2−3{\displaystyle 2-{\sqrt {3}}}
12−3{\displaystyle {\frac {1}{2-{\sqrt {3}}}}}
12−3⋅2+32+3{\displaystyle {\frac {1}{2-{\sqrt {3}}}}\cdot {\frac {2+{\sqrt {3}}}{2+{\sqrt {3}}}}}
12−3⋅2+32+3=2+34−3=2+3{\displaystyle {\frac {1}{2-{\sqrt {3}}}}\cdot {\frac {2+{\sqrt {3}}}{2+{\sqrt {3}}}}={\frac {2+{\sqrt {3}}}{4-3}}=2+{\sqrt {3}}} Do not be thrown off by the fact that the reciprocal is the conjugate. This is just a coincidence.
333{\displaystyle {\frac {3}{\sqrt[{3}]{3}}}}
331/3{\displaystyle {\frac {3}{3^{1/3}}}}
331/3⋅32/332/3{\displaystyle {\frac {3}{3^{1/3}}}\cdot {\frac {3^{2/3}}{3^{2/3}}}} This can generalize to nth roots in the denominator. If we have 1a1/n,{\displaystyle {\frac {1}{a^{1/n}}},} we multiply the top and bottom by a1−1n. {\displaystyle a^{1-{\frac {1}{n}}}. } This will make the exponent in the denominator 1.
331/3⋅32/332/3=32/3{\displaystyle {\frac {3}{3^{1/3}}}\cdot {\frac {3^{2/3}}{3^{2/3}}}=3^{2/3}} If you need to write it in radical form, factor out the 1/3. {\displaystyle 1/3. } 32/3=(32)1/3=93{\displaystyle 3^{2/3}=(3^{2})^{1/3}={\sqrt[{3}]{9}}}
