Rationalizarea numitorului unei fractii

Dupa ce am invatat sa scoatem factorii de sub radicali sau sa introducem factorii sub radicali, acum o sa invatam Rationalizarea numitorului unei fractii .

Operatia de rationalizare a numitorilor unei fractii exprimata printr-un numar irational de forma a\sqrt{b} sau a\sqrt{b}\pm c\sqrt{d}, a, c\in Q^{*} si b, d\in Q_{+}^{*}  este operatia in urma careia, prin amplificarea fractiei cu un factor, numitorul obtinut se transforma intr-un numar irational.

Deosebim urmatoarele cazuri:

  •  Rationalizarea numitorilor de forma: a\sqrt{b}, a\in Q, b\in Q_{+}^{*} intr-un astfel de caz procedam astfel \frac{c}{a\sqrt{b}}=\frac{c\cdot \sqrt{b}}{\left(\sqrt{b}\right)^{2}}=\frac{c\sqrt{b}}{a\cdot b}, a\in Q, b\in Q_{+}^{*}

Exmplu: \frac{15}{\sqrt{5}}=\frac{15\cdot\sqrt{5}}{\left(\sqrt{5}\right)^{2}}=\frac{15\sqrt{5}}{5}=\frac{3\sqrt{5}}{1}=3\sqrt{5}

Dupa ce am rationalizat numitorul am simplificat fractia prin 5 si astfel am obtinut rezultatul.

  • Rationalizarea numitorilor de forma a\sqrt{b}\pm c\sqrt{d}, a, c\in Q^{*}, b, d\in Q_{+}^{*}

In acest caz folosim formula \left(a+b\right)\cdot\left(a-b\right)=a^{2}-b^{2}, ca sa rationalizam numitorii de forma a\sqrt{b}\pm c\sqrt{d}, a, c\in Q^{*}, b, d\in Q_{+}^{*} se face prin amplificarea fractiei cu $latex a\sqrt{b} \mp c\sqrt{d}$, iar dupa amplificare numitorul devine un numar rational.

Exemplu:

1) Calculati:

a) \left(\frac{1}{\sqrt{6}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{2}}\right):\left(\sqrt{3}-1\right) =\left(\frac{1\left(\sqrt{6}-\sqrt{5}\right)}{\left(\sqrt{6}\right)^{2}-\left(\sqrt{5}\right)^{2}}+ \frac{1\left(\sqrt{5}-\sqrt{4}\right)}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{4}\right)^{2}}+ \frac{1\left(\sqrt{4}-\sqrt{3}\right)}{\left(\sqrt{4}\right)^{2}-\left(\sqrt{3}\right)^{2}}+ \frac{1\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}\right):\left(\sqrt{3}-1\right)= \left(\frac{\sqrt{6}-\sqrt{5}}{6-5}+\frac{\sqrt{5}-\sqrt{4}}{5-4}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+\frac{\sqrt{3}-\sqrt{2}}{3-2}\right):\left(\sqrt{3}-1\right) =\left(\frac{\sqrt{6}-\sqrt{5}}{1}+\frac{\sqrt{5}-\sqrt{4}}{1}+\frac{\sqrt{4}-\sqrt{3}}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}\right):\left(\sqrt{3}-1\right)= \left(\sqrt{6}-\sqrt{2}\right):\left(\sqrt{3}-1\right)= \frac{\sqrt{2}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)}= \sqrt{2}.

Ca sa rezolvam exercitiul de mai sus prima data am rationalizat numitorii asa cum am invatat mai sus, iar apoi am efectuat calculele. Am dat  factor comun ca sa putem simplifica si astfel am obtinut rezultatul.

b) \left(\frac{6}{3\sqrt{2}+2\sqrt{3}}+\frac{3}{2\sqrt{3}+3}\right):\left(\sqrt{2}-1\right)=

\left(\frac{6\left(3\sqrt{2}-2\sqrt{3}\right)}{\left(3\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}+\frac{3\left(2\sqrt{3}-3\right)}{\left(2\sqrt{3}\right)^{2}-3^{2}}\right):\left(\sqrt{2}-1\right)=

\left(\frac{18\sqrt{2}-12\sqrt{3}}{9\cdot 2-4\cdot 3}+\frac{6\sqrt{3}-9}{4\cdot 3-9}\right):\left(\sqrt{2}-1\right)=\left(\frac{18\sqrt{2}-12\sqrt{3}}{18-12}+\frac{6\sqrt{3}-9}{12-9}\right):\left(\sqrt{2}-1\right)=

\left(\frac{6\left(3\sqrt{2}-2\sqrt{3}\right)}{6}+\frac{3\left(2\sqrt{3}-3\right)}{3}\right):\left(\sqrt{2}-1\right)= \left(3\sqrt{2}-2\sqrt{3}+2\sqrt{3}-3\right):\left(\sqrt{2}-1\right)= \left(3\sqrt{2}-3\right):\left(\sqrt{2}-1\right)=\frac{3\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=3

La fel ca si la exercitiul de mai sus, prima data am rationalizat numitorii, apoi am dat factor comun pentru a putea simplifica numaratorul cu numitorul, apoi am efectuat calculele.

Apoi am dat iar factor comun pentru a simplifica rezultatul pe care l-am gasit in paranteza cu restul si astfel am gasit rezultatul 3.

c) \left(\frac{5}{\sqrt{18}}+\frac{3}{4\sqrt{2}}-\frac{7}{\sqrt{72}}\right):\frac{15}{8\sqrt{2}}=  \left(\frac{5\sqrt{18}}{18}+\frac{3\sqrt{2}}{4\cdot 2}-\frac{7\sqrt{72}}{72}\right):\frac{15\sqrt{2}}{8\cdot 2}=  \left(\frac{5\cdot 3\sqrt{2}}{18}+\frac{3\sqrt{2}}{8}-\frac{7\cdot 6\sqrt{2}}{72}\right):\frac{15\sqrt{2}}{16}=  \left(\frac{5\sqrt{2}}{6}+\frac{3\sqrt{2}}{8}-\frac{7\sqrt{2}}{12}\right)\cdot\frac{16}{15\sqrt{2}}=  \left(\frac{4\cdot 5\sqrt{2}+3\cdot 3\sqrt{2}-2\cdot 7\sqrt{2}}{24}\right)\cdot \frac{16}{15\sqrt{2}}=  \left(\frac{20\sqrt{2}+9\sqrt{2}-14\sqrt{2}}{24}\right)\cdot\frac{16}{15\sqrt{2}}=  \frac{15\sqrt{2}}{24}\cdot\frac{16}{15\sqrt{2}}=\frac{1}{2}

Foarte important! La rationalizarea numitorilor trebuie sa intelegem regulile de rationalizare, dar si scoaterea factorilor de sub radicali cat si introducerea factorilor sub radicali.

4 păreri la “Rationalizarea numitorului unei fractii

  1. Pingback: Maillot Arsenal Pas Cher

  2. Pingback: Peruvian deep wave

  3. Pingback: Smart Balance Wheel

  4. Pingback: 惡搞新聞

Lasă un răspuns