Home/ Intrebari/Q 83183
Urmator
In Process

AniDeȘcoală.ro Latest Intrebari

GabyX
Pe: 2013-10-06T09:22:12+03:00 In: In:

rationalizarea numitorului

sa se rationalizeze numitorii fractiilor:
1)1/radical de ordinul3 din 4 + radical din oridunl 3 din 10 + radical din ordinul 3 din 25
2)1/radical de ordinul 3 din 3 +radical din 2
3)1/radical de ordinul 4 din 7 – radical din ordinul 4 din 3

Similare

1 raspuns

  1. gunty
    maestru (V)

        \[ \begin{array}{l} \frac{1}{{\sqrt[3]{4} + \sqrt[3]{{10}} + \sqrt[3]{{25}}}} = \frac{{\sqrt[3]{5} - \sqrt[3]{2}}}{{\left( {\sqrt[3]{5} - \sqrt[3]{2}} \right)\left( {\sqrt[3]{4} + \sqrt[3]{{10}} + \sqrt[3]{{25}}} \right)}} = \frac{{\sqrt[3]{5} - \sqrt[3]{2}}}{3}. \\ \frac{1}{{\sqrt[3]{3} + \sqrt 2 }} = \frac{{\sqrt[3]{9} - \sqrt[3]{3} \cdot \sqrt 2 + 2}}{{\left( {\sqrt[3]{3} + \sqrt 2 } \right)\left( {\sqrt[3]{9} - \sqrt[3]{3} \cdot \sqrt 2 + 2} \right)}} = \frac{{\sqrt[3]{9} - \sqrt[6]{{72}} + 2}}{{3 + 2\sqrt 2 }} = \frac{{\left( {\sqrt[3]{9} - \sqrt[6]{{72}} + 2} \right)\left( {3 - 2\sqrt 2 } \right)}}{{\left( {3 + 2\sqrt 2 } \right)\left( {3 - 2\sqrt 2 } \right)}} = \left( {\sqrt[3]{9} - \sqrt[6]{{72}} + 2} \right)\left( {3 - 2\sqrt 2 } \right). \\ \frac{1}{{\sqrt[4]{7} - \sqrt[4]{3}}} = \frac{{\sqrt[4]{7} + \sqrt[4]{3}}}{{\left( {\sqrt[4]{7} - \sqrt[4]{3}} \right)\left( {\sqrt[4]{7} + \sqrt[4]{3}} \right)}} = \frac{{\sqrt[4]{7} + \sqrt[4]{3}}}{{\sqrt 7 - \sqrt 3 }} = \frac{{\left( {\sqrt[4]{7} + \sqrt[4]{3}} \right)\left( {\sqrt 7 + \sqrt 3 } \right)}}{{\left( {\sqrt 7 - \sqrt 3 } \right)\left( {\sqrt 7 + \sqrt 3 } \right)}} = \frac{{\left( {\sqrt[4]{7} + \sqrt[4]{3}} \right)\left( {\sqrt 7 + \sqrt 3 } \right)}}{4}. \\ \end{array} \]

Raspunde

Raspunde