Cho biểu thức $M = \frac{{{x^2} + \sqrt x }}{{x - \sqrt x + 1}} + 1 - \frac{{2x + \sqrt x }}{{\sqrt x }}$.

a) Tìm ĐKXĐ.

b) Rút gọn M.

c) Tính giá trị của M với $x = 3 - 2\sqrt 2$.

d) Tìm x để M = 2.

Lời giải

a) ĐKXĐ: $\left\{ \begin{array}{l}x \ge 0\\\sqrt x \ne 0\\x - \sqrt x + 1 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\x \ne 0\\{\left( {\sqrt x - \frac{1}{2}} \right)^2} + \frac{3}{4} \ne 0\end{array} \right. \Leftrightarrow x > 0$.

b) Ta có $M = \frac{{{x^2} + \sqrt x }}{{x - \sqrt x + 1}} + 1 - \frac{{2x + \sqrt x }}{{\sqrt x }}$

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

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

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

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

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

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

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

$= \sqrt x \left( {\sqrt x - 1} \right) = x - \sqrt x$.

c) Với $x = 3 - 2\sqrt 2$ thì

$M = 3 - 2\sqrt 2 - \sqrt {3 - 2\sqrt 2 }$

$= 3 - 2\sqrt 2 - \sqrt {2 - 2\sqrt 2 + 1}$

$= 3 - 2\sqrt 2 - \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}}$

$= 3 - 2\sqrt 2 - \sqrt 2 + 1$$= 4 - 3\sqrt 2$.

d) Để M = 2 thì $x - \sqrt x = 2$

$\Leftrightarrow x - \sqrt x - 2 = 0$

$\Leftrightarrow x - 2\sqrt x + \sqrt x - 2 = 0$

$\Leftrightarrow \sqrt x \left( {\sqrt x - 2} \right) + \sqrt x - 2 = 0$

$\Leftrightarrow \left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right) = 0$

$\Rightarrow \sqrt x - 2 = 0$ (vì $\sqrt x + 1 > 0\,\,\forall x \ge 0$)

$\Leftrightarrow \sqrt x = 2 \Rightarrow x = 4$.