Cho biểu thức: \(P = \frac{{2\sqrt x }}{{x - 1}} - \frac{1}{{1 - \sqrt x }} + \frac{{\sqrt x }}{{\sqrt x + 1}}\) với \(x \ge 0;x \ne 1\).
a) Rút gọn \(P.\)
b) Tìm \(x\) để \(P = 3\).
A.\(\begin{array}{l}{\rm{a)}}\,\,P = \frac{{\sqrt x - 1}}{{\sqrt x + 1}}\\{\rm{b)}}\,\,x = 1\end{array}\)
B.\(\begin{array}{l}{\rm{a)}}\,\,P = \frac{1}{{\sqrt x - 1}}\\{\rm{b)}}\,\,x = 2\end{array}\)
C.\(\begin{array}{l}{\rm{a)}}\,\,P = \frac{1}{{\sqrt x + 1}}\\{\rm{b)}}\,\,x = 3\end{array}\)
D.\(\begin{array}{l}{\rm{a)}}\,\,P = \frac{{\sqrt x + 1}}{{\sqrt x - 1}}\\{\rm{b)}}\,\,x = 4\end{array}\)
a) Rút gọn \(P.\)
b) Tìm \(x\) để \(P = 3\).
A.\(\begin{array}{l}{\rm{a)}}\,\,P = \frac{{\sqrt x - 1}}{{\sqrt x + 1}}\\{\rm{b)}}\,\,x = 1\end{array}\)
B.\(\begin{array}{l}{\rm{a)}}\,\,P = \frac{1}{{\sqrt x - 1}}\\{\rm{b)}}\,\,x = 2\end{array}\)
C.\(\begin{array}{l}{\rm{a)}}\,\,P = \frac{1}{{\sqrt x + 1}}\\{\rm{b)}}\,\,x = 3\end{array}\)
D.\(\begin{array}{l}{\rm{a)}}\,\,P = \frac{{\sqrt x + 1}}{{\sqrt x - 1}}\\{\rm{b)}}\,\,x = 4\end{array}\)
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Hóa Học lớp 12
