Cho \(\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}}{{{(2\tan x+\cot x)}^{2}}}dx=a+b\sqrt{3}+c\pi (*)\).
Biết rằng, tồn tại duy nhất bộ ba số hữu tỉ a, b, c thỏa mãn (*). Tổng \(a+b+c\) có giá trị bằng bao nhiêu (nhập đáp án vào ô trống, kết quả làm tròn đến chữ số thập phân thứ hai)?
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Phương pháp giải
Đưa về tích phân của các hàm cơ bản.
Lời giải
\({{(2\tan x+\cot x)}^{2}}=4{{\tan }^{2}}x+4\tan x\cot x+{{\cot }^{2}}x=\frac{4}{{{\cos }^{2}}x}+\frac{1}{{{\sin }^{2}}x}-1\).
\(\begin{array}{*{35}{l}} \int_{\frac{\pi }{6}}^{\frac{\pi }{4}}{{{(2\tan x+\cot x)}^{2}}}dx & =\int_{\frac{\pi }{6}}^{\frac{\pi }{4}}{\left( \frac{4}{{{\cos }^{2}}x}+\frac{1}{{{\sin }^{2}}x}-1 \right)}dx \\ {} & =\left. (4\tan x-\cot x-x) \right|_{\frac{\pi }{6}}^{\frac{\pi }{4}} \\ {} & =\left( 3-\frac{\pi }{4} \right)-\left( \frac{\sqrt{3}}{3}-\frac{\pi }{6} \right) \\ {} & =3-\frac{1}{3}\cdot \sqrt{3}-\frac{1}{12}\pi . \\\end{array}\)
\(\Rightarrow a+b+c=3-\frac{1}{3}-\frac{1}{12}=\frac{31}{12}\).
Đề Thi Minh Họa Đánh Giá Năng Lực Năm 2025 – ĐHQG Hà Nội – Đề Số 1 là bài thi được thiết kế hiện đại, đánh giá toàn diện ba nhóm năng lực cốt lõi: Giải Quyết Vấn Đề Và Sáng Tạo, Giao Tiếp Và Hợp Tác, Tự Chủ Và Tự Học. Với cấu trúc ba phần rõ ràng, thí sinh sẽ trải qua Toán Học Và Xử Lí Số Liệu/Tư Duy Định Lượng, Văn Học - Ngôn Ngữ/Tư Duy Định Tính và Khoa Học/Tiếng Anh. Từng câu hỏi trong đề thi không chỉ đánh giá kiến thức mà còn đo lường khả năng lập luận, tư duy logic, ngôn ngữ, tính toán, và cả kỹ năng tiếng Anh. Đây là cơ hội để học sinh thể hiện toàn diện năng lực học tập và khả năng phân tích, giải quyết vấn đề một cách khoa học và sáng tạo.
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1943
2010
2015
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13,4
14,1
14,7
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14,3
10,3
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10,1
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0
3,1
3,9
4,6
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(Nhấp vào ô để chọn đúng / sai)
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Read the following passage about food addiction and mark the letter A, B, C, or D on your answer sheet to indicate the best answer to each of the following questions from 31 to 40.
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(Adapted from Friends Global)
Which of the following best summarises the passage?
Đâu là nguyên nhân gây bệnh bên trong?