Cho biểu thức: \(P = \frac{{2{x^2} - 1}}{{{x^2} +...
Câu hỏi: Cho biểu thức: \(P = \frac{{2{x^2} - 1}}{{{x^2} + x}} - \frac{{x - 1}}{x} + \frac{3}{{x + 1}}\)1. Rút gọn \(P\) .2. Tìm x để \(P = 0\)3. Tính giá trị biểu thức \(P\) khi \(x\) thỏa mãn: \({x^2} - x = 0\).4. Tìm giá trị lớn nhất của biểu thức \(Q = \frac{1}{{{x^2} - 9}}.P\)
A \(\begin{array}{l}1)\,\,P = \frac{{x + 3}}{{x + 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,P = 2\\2)\,\,x = - 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,maxQ = \frac{{ - 1}}{4}\end{array}\)
B \(\begin{array}{l}1)\,\,P = \frac{{x + 3}}{{x + 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,P = 2\\2)\,\,x = 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,maxQ = \frac{1}{4}\end{array}\)
C \(\begin{array}{l}1)\,\,P = \frac{{x - 3}}{{x - 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,P = - 2\\2)\,\,x = - 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,maxQ = \frac{{ - 1}}{4}\end{array}\)
D \(\begin{array}{l}1)\,\,P = \frac{{x - 3}}{{x - 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,P = - 2\\2)\,\,x = - 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,maxQ = \frac{1}{4}\end{array}\)
Câu hỏi trên thuộc đề trắc nghiệm
Đề thi HK1 Toán 8 - Trường Tạ Quang Bửu - Hà Nội - Năm 2017 - 2018 (có giải chi tiết).