Question

Trục căn thức ở mẫu:

a) \(\dfrac{x^2+x}{\sqrt{x+1}}\) với x > - 1;                       b) \(\dfrac{3}{\sqrt{x}-2}\) với x > 0, x ≠ 4;

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

Answer 1

a. \(\frac{{x_{}^2 + x}}{{\sqrt {x + 1} }} = \frac{{x\left( {x + 1} \right)\sqrt {x + 1} }}{{\sqrt {x + 1} .\sqrt {x + 1} }} = \frac{{x\left( {x + 1} \right)\sqrt {x + 1} }}{{x + 1}} = x\sqrt {x + 1} \).

b. \(\frac{3}{{\sqrt x  - 2}} = \frac{{3\left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}} = \frac{{3\left( {\sqrt x  + 2} \right)}}{{x - 4}}\).

c. \(\frac{{\sqrt 3  - \sqrt 5 }}{{\sqrt 3  + \sqrt 5 }} = \frac{{\left( {\sqrt 3  - \sqrt 5 } \right)\left( {\sqrt 3  - \sqrt 5 } \right)}}{{\left( {\sqrt 3  + \sqrt 5 } \right)\left( {\sqrt 3  - \sqrt 5 } \right)}}\)

\( = \frac{{3 - 2\sqrt{15}  + 5}}{{3 - 5}} = \frac{{8 - 2\sqrt {15} }}{{ - 2}} = \frac{{ - 2\left( { - 4 + \sqrt {15} } \right)}}{{ - 2}} =  - 4 + \sqrt{15} \).

d. \(\frac{{x_{}^2 - 9}}{{\sqrt x  - \sqrt 3 }} = \frac{{\left( {x - 3} \right)\left( {x + 3} \right)\left( {\sqrt x  + \sqrt 3 } \right)}}{{\left( {\sqrt x  - \sqrt 3 } \right)\left( {\sqrt x  + \sqrt 3 } \right)}}\)

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