Đáp án:
\(\begin{array}{l}
2)\dfrac{{3\sqrt x }}{{\sqrt x - 3}}\\
4)\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
6)\dfrac{{\sqrt x + 1}}{{\sqrt x + 3}}\\
8)\dfrac{{ - 5}}{{\sqrt x - 3}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2)DK:x \ge 0;x \ne 9\\
A = \dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \left( {\sqrt x + 1} \right)\left( {\sqrt x + 3} \right) + 11\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2x - 6\sqrt x + x + 4\sqrt x + 3 + 11\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{3x + 9\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{3\sqrt x }}{{\sqrt x - 3}}\\
4)DK:x \ge 0;x \ne 1\\
A = \dfrac{{\sqrt x \left( {\sqrt x + 1} \right) + 3\left( {\sqrt x - 1} \right) - 6\sqrt x + 4}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x + \sqrt x + 3\sqrt x - 3 - 6\sqrt x + 4}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
6)DK:x \ge 0;x \ne 9\\
A = \dfrac{{\sqrt x \left( {\sqrt x + 3} \right) - x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{3}\\
= \dfrac{{x + 3\sqrt x - x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{3}\\
= \dfrac{{3\sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{3}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x + 3}}\\
8)DK:x \ge 0;x \ne 9\\
A = \dfrac{{\sqrt x \left( {\sqrt x - 3} \right) - 2\left( {\sqrt x + 3} \right) - x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 3\sqrt x - 2\sqrt x - 6 - x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - 5\sqrt x - 15}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - 5}}{{\sqrt x - 3}}
\end{array}\)
( câu 10 xem lại đề nha )
