Trần Đăng Quang

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(Thường được cập nhật sau 1 giờ!)

Bước 1: Tính toán vector \(\vec{BI}\) Theo để bài và công thức đã cho, ta có thể biểu diễn vector \(\vec{BI}\) qua các vector cơ sở \(\vec{BA}\) và \(\vec{BC}\):\(\vec{BI}=\frac{1}{2}(\vec{BA}+\vec{BM})=\frac{1}{2}(\vec{BA}+\frac{1}{2}\vec{BC})=\frac{1}{2}\vec{BA}+\frac{1}{4}\vec{BC}\)Bước 2: Tính toán vector \(\vec{BK}\) Ta biểu diễn vector \(\vec{BK}\) qua các vector \(\vec{BA}\) và \(\vec{BC}\).Sử dụng quy tắc cộng vector, ta có \(\vec{BK}=\vec{BA}+\vec{AK}\).Theo giả thiết \(AK=\frac{1}{3}AC\), nên \(\vec{AK}=\frac{1}{3}\vec{AC}\).Ta cũng có \(\vec{AC}=\vec{BC}-\vec{BA}\).Thay vào biểu thức của \(\vec{BK}\):\(\vec{BK}=\vec{BA}+\frac{1}{3}\vec{AC}=\vec{BA}+\frac{1}{3}(\vec{BC}-\vec{BA})=\vec{BA}+\frac{1}{3}\vec{BC}-\frac{1}{3}\vec{BA}=\frac{2}{3}\vec{BA}+\frac{1}{3}\vec{BC}\)Bước 3: So sánh hai vector và kết luận So sánh hai vector \(\vec{BI}\) và \(\vec{BK}\):\(\vec{BI}=\frac{1}{2}\vec{BA}+\frac{1}{4}\vec{BC}\)\(\vec{BK}=\frac{2}{3}\vec{BA}+\frac{1}{3}\vec{BC}\)Ta thấy rằng \(\vec{BK}=\frac{4}{3}(\frac{1}{2}\vec{BA}+\frac{1}{4}\vec{BC})\). Do đó:\(\vec{BK}=\frac{4}{3}\vec{BI}\)Vì vector \(\vec{BK}\) tỉ lệ với vector \(\vec{BI}\) (hệ số tỉ lệ \(k=\frac{4}{3}\)), và cả hai vector đều xuất phát từ điểm chung B, nên ba điểm B, I, K thẳng hàng. Trả lời: Ba điểm B, I, K thẳng hàng.  Đang tạo đường liên kết công khai...Cảm ơn bạnÝ kiến phản hồi của bạn sẽ giúp Google cải thiện. 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Sử dụng quy tắc chèn điểm theo một cách khác:\(\vec{MN}=\vec{MA}+\vec{AC}+\vec{CN}\)\(\vec{MN}=\vec{MD}+\vec{DB}+\vec{BN}\)Cộng hai phương trình này lại:\(2\vec{MN}=(\vec{MA}+\vec{MD})+(\vec{AC}+\vec{DB})+(\vec{CN}+\vec{BN})\)Với \(\vec{MA}+\vec{MD}=\vec{0}\) và \(\vec{CN}+\vec{BN}=\vec{0}\), ta có:\(2\vec{MN}=\vec{0}+\vec{AC}+\vec{DB}+\vec{0}\)\(\vec{MN}=\frac{1}{2}(\vec{AC}+\vec{DB})\)Từ kết quả ở Bước 1 và Bước 2, ta suy ra \(\vec{MN}=\frac{1}{2}(\vec{AB}+\vec{DC})=\frac{1}{2}(\vec{AC}+\vec{DB})\). Bước 3: Chứng minh câu b) Ta có \(I\) là trung điểm của \(MN\). Áp dụng tính chất trung điểm cho các đoạn thẳng \(AD,BC\) và \(MN\) đối với điểm \(I\):\(\vec{IA}+\vec{ID}=2\vec{IM}\)\(\vec{IB}+\vec{IC}=2\vec{IN}\)Cộng các phương trình trên, ta được tổng vectơ cần tính:\(\vec{IA}+\vec{IB}+\vec{IC}+\vec{ID}=(\vec{IA}+\vec{ID})+(\vec{IB}+\vec{IC})=2\vec{IM}+2\vec{IN}=2(\vec{IM}+\vec{IN})\)Vì \(I\) là trung điểm của \(MN\), theo tính chất trung điểm, \(\vec{IM}+\vec{IN}=\vec{0}\).Do đó:\(\vec{IA}+\vec{IB}+\vec{IC}+\vec{ID}=2(\vec{0})=\vec{0}\)Trả lời: Các đẳng thức đã được chứng minh trong các bước trên. Cụ thể:a) \(\vec{MN}=\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}\vec{\mathbf{AB}}\mathbf{+}\vec{\mathbf{DC}}\mathbf{)=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}\vec{\mathbf{AC}}\mathbf{+}\vec{\mathbf{DB}}\mathbf{)}\)b) \(\vec{IA}+\vec{IB}+\vec{IC}+\vec{ID}=\vec{\mathbf{0}}\)  Đang tạo đường liên kết công khai...Cảm ơn bạnÝ kiến phản hồi của bạn sẽ giúp Google cải thiện. 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ta biểu diễn các vectơ dưới dạng hiệu:\(\vec{AC}=\vec{C}-\vec{A}\)\(\vec{BD}=\vec{D}-\vec{B}\)\(\vec{AC}+\vec{BD}=(\vec{C}-\vec{A})+(\vec{D}-\vec{B})=\vec{C}+\vec{D}-\vec{A}-\vec{B}\)Và \(2\vec{EF}=2(\vec{F}-\vec{E})=2(\frac{\vec{C}+\vec{D}}{2}-\frac{\vec{A}+\vec{B}}{2})=2(\frac{\vec{C}+\vec{D}-\vec{A}-\vec{B}}{2})=\vec{C}+\vec{D}-\vec{A}-\vec{B}\)Do đó, \(\vec{AC}+\vec{BD}=2\vec{EF}\). Tương tự, chứng minh \(\vec{AD}+\vec{BC}=2\vec{EF}\):\(\vec{AD}=\vec{D}-\vec{A}\)\(\vec{BC}=\vec{C}-\vec{B}\)\(\vec{AD}+\vec{BC}=(\vec{D}-\vec{A})+(\vec{C}-\vec{B})=\vec{C}+\vec{D}-\vec{A}-\vec{B}\)Do đó, \(\vec{AD}+\vec{BC}=2\vec{EF}\). Từ hai đẳng thức trên, ta có \(\vec{AC}+\vec{BD}=\vec{AD}+\vec{BC}=2\vec{EF}\).

Nghi ngờ của Bình là hoàn toàn chính xác. Giá trị \(0,290\) là sai số thô và cần được loại bỏ trước khi tính toán kết quả cuối cùng để đảm bảo độ tin cậy của phép đo rơi tự do.

Độ lệch chuẩn càng nhỏ thì mức độ phân tán của số liệu càng thấp. Do đó, vận động viên A có thành tích bắn thử ổn định hơn vận động viên B.


Khoảng biến thiên của mẫu số liệu trên là 10. Không nên dùng đại lượng này để đo độ phân tán của mẫu số liệu trên vì nó rất nhạy cảm với các giá trị ngoại lệ

Hà nội :12,5°C

TP HỒ CHÍ MINH:HN: 3,2°C 

NHẬN XÉT :Nhiệt độ trung bình các tháng trong năm tại Hà Nội biến động mạnh hơn đáng kể so với Thành phố Hồ Chí Minh. Các chỉ số khoảng biến thiên, khoảng tứ phân vị và độ lệch chuẩn của Hà Nội đều cao hơn rất nhiều so với TP. Hồ Chí Minh. Điều này cho thấy khí hậu Hà Nội có sự phân hóa mùa rõ rệt (mùa đông lạnh, mùa hè nóng), trong khi khí hậu TP. Hồ Chí Minh ổn định hơn, nhiệt độ các tháng ít chênh lệch.