\(K = \dfrac{{\sqrt x }}{{2\left( {\sqrt x + 3} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l} DK:x > 0;x \ne 9\\ K = \left( {\dfrac{1}{{\sqrt x + 3}} – \dfrac{5}{{9 – x}}} \right):\left( {\dfrac{{3\sqrt x + 1}}{{x – 3\sqrt x }} – \dfrac{1}{{\sqrt x }}} \right)\\ = \left[ {\dfrac{{\sqrt x – 3 + 5}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}} \right]:\left[ {\dfrac{{3\sqrt x + 1 – \sqrt x + 3}}{{\sqrt x \left( {\sqrt x – 3} \right)}}} \right]\\ = \dfrac{{\sqrt x + 2}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x – 3} \right)}}{{2\sqrt x + 4}}\\ = \dfrac{{\sqrt x + 2}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x – 3} \right)}}{{2\left( {\sqrt x + 2} \right)}}\\ = \dfrac{{\sqrt x }}{{2\left( {\sqrt x + 3} \right)}} \end{array}\)
Đáp án:
Giải thích các bước giải:
Đáp án:
\(K = \dfrac{{\sqrt x }}{{2\left( {\sqrt x + 3} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x > 0;x \ne 9\\
K = \left( {\dfrac{1}{{\sqrt x + 3}} – \dfrac{5}{{9 – x}}} \right):\left( {\dfrac{{3\sqrt x + 1}}{{x – 3\sqrt x }} – \dfrac{1}{{\sqrt x }}} \right)\\
= \left[ {\dfrac{{\sqrt x – 3 + 5}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}} \right]:\left[ {\dfrac{{3\sqrt x + 1 – \sqrt x + 3}}{{\sqrt x \left( {\sqrt x – 3} \right)}}} \right]\\
= \dfrac{{\sqrt x + 2}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x – 3} \right)}}{{2\sqrt x + 4}}\\
= \dfrac{{\sqrt x + 2}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x – 3} \right)}}{{2\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x }}{{2\left( {\sqrt x + 3} \right)}}
\end{array}\)