There are several types of irrationality of fraction in a denominator. It is connected with presence at it of an algebraic root of one or various degrees. To get rid of irrationality, it is necessary to perform certain mathematical operations depending on a situation.
Instruction
1. Before getting rid of irrationality of fraction in a denominator, it is necessary to define its type, and depending on it to continue the decision. And though any irrationality follows from simple presence of roots, their various combinations and degrees assume different algorithms.
2. Square root in a denominator, expression of a type of a / √ bvvedite the additional multiplier equal to √b. That the fraction did not change, it is necessary to multiply both numerator, and a denominator: a/√ b → (· b)/b. Example 1: 10/√ 3 → (10•√3)/3.
3. Below the line split up existence a root of fractional degree of a look m/n, and n> meto expression looks as follows: a/√ (b^m/n).
4. Get rid of similar irrationality also by input of a multiplier, more difficult this time: b^(n-m)/n, i.e. it is necessary to subtract extent of expression under its sign from an exponent of the root. Then in a denominator there will be the first degree: a/(b^m/n) → to a • √ (b^(n-m) / n)/b. Example 2: 5/(4^3/5) → 5 • √ (4^2/5)/4 = 5 • √ (16^1/5)/4.
5. The sum square to a korneyumnozhta both making fractions on a similar difference. Then from irrational addition of roots the denominator will be transformed to the difference of expressions/numbers under the sign of a root: a/(√ b + √c) → a • (√b - √c) / (b - c). Example 3: 9/(√ 13 + √23) → 9 • (√13 - √23) / (13 - 23) = 9 · (√23 - √13)/10.
6. A sum/difference cubic you korneyvyberit an incomplete square of a difference as an additional multiplier if in a denominator there is a sum, and respectively an incomplete square of the sum for the difference of roots: a/(∛ b ± ∛c) → a • (∛b² ∓ ∛ (b•c) + ∛c²)/((∛b ± ∛c) • ∛b² ∓ ∛ (b•c) + ∛c²) →a • (∛b² ∓ ∛ (b•c) + ∛c²) / (b ± c). Example 4: 7/(∛ 5 + ∛4) → 7 • (∛25-∛20 + ∛16)/9.
7. If at a task there is both a square and cubic root, then divide the decision into two stages: consistently bring out of a denominator a square root, and then cubic. It becomes by methods already known to you: in the first action it is necessary to choose a multiplier of a difference/sum of roots, in the second – an incomplete square of a sum/difference.