P= ( căn x+2/x – 5 căn x +6 – căn x+2/ 2- căn x- căn x+2/căn x-3) : (2- căn x/ căn x+1)
a) Rút gọn P
b)Tìm x để 1/P ≤ -5/2
P= ( căn x+2/x – 5 căn x +6 – căn x+2/ 2- căn x- căn x+2/căn x-3) : (2- căn x/ căn x+1)
a) Rút gọn P
b)Tìm x để 1/P ≤ -5/2
Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ge 0\\
x \ne 4\\
x \ne 9
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
a,\\
P = \left( {\dfrac{{\sqrt x + 2}}{{x – 5\sqrt x + 6}} – \dfrac{{\sqrt x + 2}}{{2 – \sqrt x }} – \dfrac{{\sqrt x + 2}}{{\sqrt x – 3}}} \right):\dfrac{{2 – \sqrt x }}{{\sqrt x + 1}}\\
= \left( {\dfrac{{\sqrt x + 2}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}} + \dfrac{{\sqrt x + 2}}{{\sqrt x – 2}} – \dfrac{{\sqrt x + 2}}{{\sqrt x – 3}}} \right):\dfrac{{2 – \sqrt x }}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 2 + \left( {\sqrt x + 2} \right).\left( {\sqrt x – 3} \right) – \left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right)}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}}:\dfrac{{2 – \sqrt x }}{{\sqrt x + 1}}\\
= \dfrac{{\left( {\sqrt x + 2} \right).\left[ {1 + \left( {\sqrt x – 3} \right) – \left( {\sqrt x – 2} \right)} \right]}}{{\left( {\sqrt x – 2} \right).\left( {\sqrt x – 3} \right)}}:\dfrac{{2 – \sqrt x }}{{\sqrt x + 1}}\\
= \dfrac{{\left( {\sqrt x + 2} \right).0}}{{\left( {\sqrt x – 2} \right).\left( {\sqrt x – 3} \right)}}:\dfrac{{2 – \sqrt x }}{{\sqrt x + 1}}\\
= 0
\end{array}\)