Cho biểu thức:

\[A = \left( {\frac{{x - 5\sqrt x }}{{x - 25}} - 1} \right):\left( {\frac{{25 - x}}{{x + 2\sqrt x - 15}} - \frac{{\sqrt x + 3}}{{\sqrt x + 5}} + \frac{{\sqrt x - 5}}{{\sqrt x - 3}}} \right)\]

a) Rút gọn A.

b) Tìm x để A < 1.

Lời giải                         

ĐKXĐ: x ≥ 0, x ≠ 25, x ≠ 9

a) \[A = \left( {\frac{{x - 5\sqrt x }}{{x - 25}} - 1} \right):\left( {\frac{{25 - x}}{{x + 2\sqrt x - 15}} - \frac{{\sqrt x + 3}}{{\sqrt x + 5}} + \frac{{\sqrt x - 5}}{{\sqrt x - 3}}} \right)\]

\[ = \frac{{x - 5\sqrt x - \left( {x - 25} \right)}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}:\left[ {\frac{{25 - x}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}} - \frac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}} + \frac{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}} \right]\]

\[ = \frac{{ - 5\left( {\sqrt x - 5} \right)}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}:\frac{{25 - x - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + x - 25}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}\]

\[ = - \frac{5}{{\sqrt x + 5}}:\frac{{ - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}\]

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

\[ = - \frac{5}{{\sqrt x + 5}} \cdot \frac{{ - \left( {\sqrt x + 5} \right)}}{{\sqrt x + 3}}\]

\[ = \frac{5}{{\sqrt x + 3}}\].

b) Với x ≥ 0, x ≠ 25, x ≠ 9, ta có

A < 1 \[ \Leftrightarrow \frac{5}{{\sqrt x + 3}} < 1\] ⇔ \[ \Leftrightarrow \frac{{5 - \left( {\sqrt x + 3} \right)}}{{\sqrt x + 3}} < 0 \Leftrightarrow \frac{{2 - \sqrt x }}{{\sqrt x + 3}} < 0\]

\( \Leftrightarrow 2 - \sqrt x < 0\) (vì \[\sqrt x + 3 > 0\] ∀x ≥ 0, x ≠ 25, x ≠ 9)

\( \Leftrightarrow \sqrt x > 2 \Leftrightarrow x > 4\)

Kết hợp điều kiện x ≥ 0, x ≠ 25, x ≠ 9 ta được: x > 4, x ≠ 25, x ≠ 9.

Vậy để A < 1 thì x > 4, x ≠ 25, x ≠ 9