Rút gọn biểu thức \(B = \frac{{2\sqrt x }}{{\sqrt x + 3}} - \frac{{\sqrt x + 1}}{{\sqrt x - 3}} - \frac{{7\sqrt x + 3}}{{9 - x}}\) (x > 0, x ≠ 9).
Hướng dẫn giải
Với x > 0, x ≠ 9, ta có:
\(B = \frac{{2\sqrt x }}{{\sqrt x + 3}} - \frac{{\sqrt x + 1}}{{\sqrt x - 3}} - \frac{{7\sqrt x + 3}}{{9 - x}}\)
\(B = \frac{{2\sqrt x \left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} - \frac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} + \frac{{7\sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\)
\(B = \frac{{2x - 6\sqrt x - x - 4\sqrt x - 3 + 7\sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\)
\(B = \frac{{x - 3\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} = \frac{{\left( {\sqrt x - 3} \right)\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} = \frac{{\sqrt x }}{{\sqrt x + 3}}\).