Question

Trực căn thức ở mẫu:

a) \(\dfrac{9}{2\sqrt{3}}\);                            b) \(\dfrac{2}{\sqrt{a}}\) với a > 0;

c) \(\dfrac{7}{3-\sqrt{2}}\);                        d) \(\dfrac{5}{\sqrt{x}+3}\) với x > 0, x ≠ 9;

e) \(\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\);                    g) \(\dfrac{1}{\sqrt{x}-\sqrt{3}}\) với x > 0, x ≠ 3.

Answer 1

a. \(\frac{9}{{2\sqrt 3 }} = \frac{{9\sqrt 3 }}{{2\sqrt 3 .\sqrt 3 }} = \frac{{9\sqrt 3 }}{{2.3}} = \frac{{9\sqrt 3 }}{6} = \frac{{3\sqrt 3 }}{2}\).

b. \(\frac{2}{{\sqrt a }} = \frac{{2\sqrt a }}{{\sqrt a .\sqrt a }} = \frac{{2\sqrt a }}{a}\).

c. \(\frac{7}{{3 - \sqrt 2 }} = \frac{{7\left( {3 + \sqrt 2 } \right)}}{{\left( {3 - \sqrt 2 } \right)\left( {3 + \sqrt 2 } \right)}} = \frac{{7\left( {3 + \sqrt 2 } \right)}}{{9 - 2}} = \frac{{7\left( {3 + \sqrt 2 } \right)}}{7} = 3 + \sqrt 2 \).

d. \(\frac{5}{{\sqrt x  + 3}} = \frac{{5\left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}} = \frac{{5\left( {\sqrt x  - 3} \right)}}{{x - 9}}\).

e. \(\frac{{\sqrt 3  - \sqrt 2 }}{{\sqrt 3  + \sqrt 2 }} = \frac{{\left( {\sqrt 3  - \sqrt 2 } \right)\left( {\sqrt 3  - \sqrt 2 } \right)}}{{\left( {\sqrt 3  + \sqrt 2 } \right)\left( {\sqrt 3  - \sqrt 2 } \right)}} = \frac{{3 - 2\sqrt 6  + 2}}{{3 - 2}} = 5 - 2\sqrt 6 \).

g. \(\frac{1}{{\sqrt x  - \sqrt 3 }} = \frac{{1\left( {\sqrt x  + \sqrt 3 } \right)}}{{\left( {\sqrt x  - \sqrt 3 } \right)\left( {\sqrt x  + \sqrt 3 } \right)}} = \frac{{\sqrt x  + \sqrt 3 }}{{x - 3}}\).