Rút gọn biểu thức: A = (x - 2 căn bậc hai x) / (x căn bậc hai x - 1) + (căn bậc hai x + 1)

Rút gọn biểu thức:

$A = \frac{{x - 2\sqrt x }}{{x\sqrt x - 1}} + \frac{{\sqrt x + 1}}{{x\sqrt x + x + \sqrt x }} + \frac{{1 + 2x - 2\sqrt x }}{{{x^2} - \sqrt x }}$ (x > 0, x ¹ 1)

$A = \frac{{x - 2\sqrt x }}{{x\sqrt x - 1}} + \frac{{\sqrt x + 1}}{{x\sqrt x + x + \sqrt x }} + \frac{{1 + 2x - 2\sqrt x }}{{{x^2} - \sqrt x }}$ (x > 0, x ¹ 1)

$= \frac{{x - 2\sqrt x }}{{x\sqrt x - 1}} + \frac{{\sqrt x + 1}}{{x\sqrt x + x + \sqrt x }} + \frac{{1 + 2x - 2\sqrt x }}{{\sqrt x \left( {x\sqrt x - 1} \right)}}$

$= \frac{{x - 2\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} + \frac{{\sqrt x + 1}}{{\sqrt x \left( {x + \sqrt x + 1} \right)}} + \frac{{1 + 2x - 2\sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}$

$= \frac{{\sqrt x \left( {x - 2\sqrt x } \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} + \frac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} + \frac{{1 + 2x - 2\sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}$

$= \frac{{x\sqrt x - 2x + x - 1 + 1 + 2x - 2\sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}$

$= \frac{{x\sqrt x + x - 2\sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}$

$= \frac{{\sqrt x \left( {x + \sqrt x - 2} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}$

$= \frac{{\sqrt x \left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}$

$= \frac{{\sqrt x + 2}}{{x + \sqrt x + 1}}$

Vậy $A = \frac{{\sqrt x + 2}}{{x + \sqrt x + 1}}$.