\(\begin{array}{l} B = \dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x + 1}}{{\sqrt x – 3}} + \dfrac{{3 – 11\sqrt x }}{{9 – x}}\\ = \dfrac{{2\sqrt x \left( {\sqrt x – 3} \right) + \left( {\sqrt x + 1} \right)\left( {\sqrt x + 3} \right) – 3 + 11\sqrt x }}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{2x – 6\sqrt x + x + 4\sqrt x + 3 – 3 + 11\sqrt x }}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{3x + 9\sqrt x }}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}} = \dfrac{{3\sqrt x }}{{\sqrt x – 3}} \end{array}\)
Mình trình bày trong hình nha
Đáp án:
\(\dfrac{{3\sqrt x }}{{\sqrt x – 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B = \dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x + 1}}{{\sqrt x – 3}} + \dfrac{{3 – 11\sqrt x }}{{9 – x}}\\
= \dfrac{{2\sqrt x \left( {\sqrt x – 3} \right) + \left( {\sqrt x + 1} \right)\left( {\sqrt x + 3} \right) – 3 + 11\sqrt x }}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2x – 6\sqrt x + x + 4\sqrt x + 3 – 3 + 11\sqrt x }}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{3x + 9\sqrt x }}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}} = \dfrac{{3\sqrt x }}{{\sqrt x – 3}}
\end{array}\)