Article in English: If your fraction contains a sum of two terms in the denominator, at least one of which is irrational, then you cannot multiply the fraction by it in the numerator and denominator. 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. The conjugate of an expression is the same expression with the sign reversed. For example, the conjugate of 2+2{\displaystyle 2+{\sqrt {2}}} is 2−2.{\displaystyle 2-{\sqrt {2}}.}  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}}}

Summary in Indonesian:
Periksalah pecahan. Kalikan pecahan dengan konjugasi penyebutnya. Sederhanakan sesuai kebutuhan.