\(\begin{array}{l} 23)\left[ {\dfrac{{\sqrt x + \sqrt x – 2 – 2\sqrt x – 4}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}} \right]:\left[ {\dfrac{{x + 2\sqrt x + 6 – x – 2\sqrt x – 4}}{{\sqrt x + 2}}} \right]\\ = \dfrac{{ – 6}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}:\dfrac{2}{{\sqrt x + 2}}\\ = \dfrac{{ – 6}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x + 2}}{2}\\ = – \dfrac{3}{{\sqrt x – 2}}\\ 24)\left[ {\dfrac{{\sqrt x + 2\sqrt x – 6 – 3\sqrt x – 9}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}} \right]:\left( {\dfrac{{x + 3\sqrt x + 12 – x – 3\sqrt x – 9}}{{\sqrt x + 3}}} \right)\\ = \dfrac{{ – 15}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x + 3}}{3}\\ = – \dfrac{5}{{\sqrt x – 3}}\\ 25)\left[ {\dfrac{{3x + 3 – \sqrt x \left( {\sqrt x + 3} \right) – 2\sqrt x \left( {\sqrt x – 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}} \right]:\left( {\dfrac{{2\sqrt x – 2 – \sqrt x + 3}}{{\sqrt x – 3}}} \right)\\ = \dfrac{{3x + 3 – x – 3\sqrt x – 2x + 6\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}:\dfrac{{\sqrt x + 1}}{{\sqrt x – 3}}\\ = \dfrac{{3\sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}.\dfrac{{\sqrt x – 3}}{{\sqrt x + 1}}\\ = \dfrac{3}{{\sqrt x + 3}}\\ 26)\left[ {\dfrac{{ – \sqrt x \left( {\sqrt x + 2} \right) – x – 4}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}} \right]:\left[ {\dfrac{{2\sqrt x + 1 – \sqrt x – 2}}{{\sqrt x \left( {\sqrt x + 2} \right)}}} \right]\\ = \dfrac{{ – x – 2\sqrt x – x – 4}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\sqrt x – 1}}\\ = \dfrac{{ – 2x – 2\sqrt x – 4}}{{\left( {\sqrt x – 2} \right)}}.\dfrac{{\sqrt x }}{{\sqrt x – 1}} \end{array}\)
Answers ( )
Đáp án:
25) \(\dfrac{3}{{\sqrt x + 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
23)\left[ {\dfrac{{\sqrt x + \sqrt x – 2 – 2\sqrt x – 4}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}} \right]:\left[ {\dfrac{{x + 2\sqrt x + 6 – x – 2\sqrt x – 4}}{{\sqrt x + 2}}} \right]\\
= \dfrac{{ – 6}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}:\dfrac{2}{{\sqrt x + 2}}\\
= \dfrac{{ – 6}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x + 2}}{2}\\
= – \dfrac{3}{{\sqrt x – 2}}\\
24)\left[ {\dfrac{{\sqrt x + 2\sqrt x – 6 – 3\sqrt x – 9}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}} \right]:\left( {\dfrac{{x + 3\sqrt x + 12 – x – 3\sqrt x – 9}}{{\sqrt x + 3}}} \right)\\
= \dfrac{{ – 15}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x + 3}}{3}\\
= – \dfrac{5}{{\sqrt x – 3}}\\
25)\left[ {\dfrac{{3x + 3 – \sqrt x \left( {\sqrt x + 3} \right) – 2\sqrt x \left( {\sqrt x – 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}} \right]:\left( {\dfrac{{2\sqrt x – 2 – \sqrt x + 3}}{{\sqrt x – 3}}} \right)\\
= \dfrac{{3x + 3 – x – 3\sqrt x – 2x + 6\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}:\dfrac{{\sqrt x + 1}}{{\sqrt x – 3}}\\
= \dfrac{{3\sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}.\dfrac{{\sqrt x – 3}}{{\sqrt x + 1}}\\
= \dfrac{3}{{\sqrt x + 3}}\\
26)\left[ {\dfrac{{ – \sqrt x \left( {\sqrt x + 2} \right) – x – 4}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}} \right]:\left[ {\dfrac{{2\sqrt x + 1 – \sqrt x – 2}}{{\sqrt x \left( {\sqrt x + 2} \right)}}} \right]\\
= \dfrac{{ – x – 2\sqrt x – x – 4}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\sqrt x – 1}}\\
= \dfrac{{ – 2x – 2\sqrt x – 4}}{{\left( {\sqrt x – 2} \right)}}.\dfrac{{\sqrt x }}{{\sqrt x – 1}}
\end{array}\)