Cho phương trình \((m-1) x^{4}+2(m-3) x^{2}+m+3=0\) ( \(m\) là tham số). Tìm \(m\) để phương trình vô nghiệm.
\(m \in(-\infty;-3) \cup\left(\frac{3}{2};+\infty\right)\).
\(m \leq-3\).
\(m>\frac{3}{2}\).
\(m<-3\).
Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Đặt \(t=x^{2},(t \geq 0)\). Khi đó ta có phương trình: \((m-1) t^{2}+2(m-3) t+m+3=0\). (1)
Với \(m=1\) thì (1) \(\Leftrightarrow-4 t+4=0 \Leftrightarrow t=1 \Leftrightarrow x= \pm 1\) (Loại)
Với \(m \neq 1\) để phương trình ban đầu vô nghiệm thì:
TH1: (1) vô nghiệm \(\Leftrightarrow \Delta^{\prime}<0 \Leftrightarrow-8 m+12<0 \Leftrightarrow m>\frac{3}{2}\).
TH2: (1) có 2 nghiệm âm
\(\Leftrightarrow\left\{\begin{array}{l}\Delta^{\prime} \geq 0 \\ \mathrm{t}_{1} \cdot \mathrm{t}_{2}>0 \\ \mathrm{t}_{1}+\mathrm{t}_{2}<0\end{array} \Leftrightarrow\left\{\begin{array}{l}-8 \mathrm{~m}+12 \geq 0 \\ \frac{\mathrm{~m}+3}{\mathrm{~m}-1}>0 \\ -\frac{2(\mathrm{~m}-3)}{\mathrm{m}+1}<0\end{array}\right.\right.\)
\(\Leftrightarrow\left\{\begin{array}{l}\mathrm{m} \leq \frac{3}{2} \\ \mathrm{~m} \in(-\infty;-3) \cup(1;+\infty) \\ \mathrm{m} \in(-\infty; 1) \cup(3;+\infty)\end{array} \Leftrightarrow \mathrm{m} \in(-\infty;-3)\right.\)
Kết hợp 2 trường hợp, ta được \(m \in(-\infty;-3) \cup\left(\frac{3}{2};+\infty\right)\).
Đề Thi Tham Khảo Đánh Giá Năng Lực Năm 2024 – ĐHQG Hà Nội – Đề Số 01 là bài kiểm tra toàn diện và khoa học, giúp học sinh thể hiện năng lực giải quyết vấn đề, tư duy sáng tạo và phân tích logic. Với ba phần thi chính: Toán Học Và Xử Lí Số Liệu, Văn Học - Ngôn Ngữ, và Khoa Học/Tiếng Anh, đề thi không chỉ dừng lại ở việc kiểm tra kiến thức cơ bản mà còn yêu cầu thí sinh phát triển khả năng lập luận và ứng dụng thực tiễn. Đặc biệt, phần thi Khoa Học cho phép lựa chọn giữa Vật Lí, Hóa Học, Sinh Học, Lịch Sử, Địa Lí tạo điều kiện cho thí sinh phát huy thế mạnh cá nhân.
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https://tvsthpt.dlib.vn/app/doc-sach-ebook/d0a01a0e4114
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