ĐỀ THI GIỮA KÌ GT3 HỌC KÌ 20193 – NHÓM NGÀNH 2 Câu 1: ∞ 1 𝑎) ∑ 3 𝑛 √ln 𝑛 𝑛=2 𝑋é𝑡 𝑓(𝑥) = . 𝐶ó 𝑢𝑛 > 0∀𝑛 ≥ 2 → 𝑐ℎ𝑢ỗ𝑖 𝑑ươ𝑛𝑔 1 3 𝑥 √ln 𝑥 1 −23 . ln 𝑥 3 ′ (𝑥) +) 𝑓 =− 2 < 0 → 𝑓(𝑥) đơ𝑛 đ𝑖ệ𝑢 𝑔𝑖ả𝑚 3 (𝑥. √ln 𝑥 ) 1 +) lim 3 =0 𝑥→∞ 𝑥 √ln 𝑥 3 2 ∞ +) ∫ 3 𝑑𝑥 = ∫ 𝑡 −3 𝑑𝑡 = 𝑡 3 = ∞ → 𝑇𝑝 𝑝ℎâ𝑛 𝑘ì 2 ln 2 2 𝑥 √ln 𝑥 ln 2 → 𝐶ℎ𝑢ỗ𝑖 đã 𝑐ℎ𝑜 𝑝ℎâ𝑛 𝑘ì 𝑡ℎ𝑒𝑜 𝑇𝐶 𝑡í𝑐ℎ 𝑝ℎâ𝑛 ∞ 1 𝑛 𝑏) ∑ √𝑛 (𝑒 2 − 1) . 𝑢𝑛 > 0∀𝑛 ≥ 1 3 𝑛=1 1 3 3 𝑢𝑛 = √𝑛 (𝑒 𝑛2 − 1) ~ √𝑛. ∞ 𝑀à ∑ 1 1 ( 𝑑𝑜 𝑒 𝑡 − 1 ~ 𝑡 𝑘ℎ𝑖 𝑡 → 0 ) 𝑛=1 𝑛 5 ℎộ𝑖 𝑡ụ > 1) → 𝐶ℎ𝑢ỗ𝑖 đã 𝑐ℎ𝑜 ℎộ𝑖 𝑡ụ 𝑡ℎ𝑒𝑜 𝑡𝑖ê𝑢 𝑐ℎ𝑢ẩ𝑛 𝑠𝑜 𝑠á𝑛ℎ Câu 2: ∞ 𝑛 2𝑥 3 ∑ ( ) √𝑛 𝑥 6 + 4 1 𝑛=1 1 1 𝑛 1 2 𝑢𝑛 (𝑥) = .( ) = 𝑛 2 . √𝑛 √𝑛 𝑥 + 4 √𝑛 𝑥 + 1 √𝑛 2 𝑋é𝑡 ∑ 𝑛 𝑙à 𝑐ℎ𝑢ỗ𝑖 𝑑ươ𝑛𝑔 𝑑𝑜 𝑛 > 0∀𝑛 ≥ 1 𝑛=1 3 2 𝑛 . √𝑛 𝑛 1 < 1 → 𝐶ℎ𝑢ỗ𝑖 ℎộ𝑖 𝑡ụ 𝑡ℎ𝑒𝑜 𝑇𝐶 𝐷′ 𝐴𝑙𝑒𝑚𝑏𝑒𝑟𝑡 𝑛→∞ 2.2𝑛 . √𝑛 + 1 2 → 𝐶ℎ𝑢ỗ𝑖 đã 𝑐ℎ𝑜 ℎộ𝑖 𝑡ụ đề𝑢 𝑡𝑟ê𝑛 𝑅 𝑡ℎ𝑒𝑜 𝑇𝐶 𝑊𝑒𝑖𝑒𝑟𝑠𝑡𝑟𝑎𝑠𝑠 Câu 3: lim = ∞ 𝑛−1 𝑛 ∑( ) (2𝑥 + 1)𝑛 2𝑛 + 1 𝑛=0 𝑋é𝑡: 1 2𝑛 + 1 1 2 lim | 𝑛 <1 | = lim | |.| |= 𝑛→∞ √𝑢 (𝑥) 𝑛→∞ 𝑛 − 1 |2𝑥 + 1| 2𝑥 + 1 𝑛 → −2 < 2𝑥 + 1 < 2 → −3 1 <𝑥<− 𝑉ậ𝑦 𝑚𝑖ề𝑛 ℎộ𝑖 𝑡ụ 𝑐ầ𝑛 𝑡ì𝑚 𝑙à (− ; − ) 2 2 Câu 4: 𝑥 𝑓(𝑥) = 𝑥𝑙𝑛(2 − 𝑥) = 𝑥. ln 2 + 𝑥. ln (1 − ) 2 𝑛+1 𝑥 ∞ (2 ) 𝑥 → 𝑓(𝑥) = 𝑥. ln 2 + 𝑥. ∑ − ∀| | < 1 𝑛+1 2 𝑛=0 ∞ 1 1 = 𝑥. ln 2 + ∑ − . 𝑛 . 𝑥 𝑛+2 ∀|𝑥| < 2 2 2 (𝑛 + 1) 𝑛=0 Câu 5: 𝑎) (1 − 𝑥) + 𝑥𝑦 ′ 𝑦 = 0 𝑑𝑦 𝑥−1 𝑦2 → (1 − 𝑥) = −𝑥 𝑦→ 𝑑𝑥 = 𝑦𝑑𝑦 → 𝑥 − ln|𝑥| + 𝐶 = 𝑑𝑥 𝑥 2 → 𝑦 = ±√2𝑥 − 2 ln|𝑥| + 𝐶 𝑏)(𝑥 − 2𝑦)𝑑𝑥 + 𝑥𝑑𝑦 = 0 +) 𝑥 = 0 𝑙à 𝑛𝑔ℎ𝑖ệ𝑚 𝑘ì 𝑑ị +) 𝑥 ≠ 0. 𝐶ó 2 𝑥 − 2𝑦 + 𝑥. 𝑦 ′ = 0 → 𝑦 ′ − 𝑦 = 1 (𝑃𝑇𝑉𝑃 𝑡𝑢𝑦ế𝑛 𝑡í𝑛ℎ ) 𝑥 2 2 ∫𝑥 𝑑𝑥 ∫ −𝑥𝑑𝑥 →𝑦=𝑒 (∫ 1. 𝑒 𝑑𝑥 + 𝐶 ) 1 = 𝑥 2 (− + 𝐶) = −𝑥 + 𝐶𝑥 2 𝑥 𝑐)(3𝑥 2 𝑦 + 2 cos 𝑦)𝑑𝑥 + (𝑥 3 − 2𝑥𝑠𝑖𝑛 𝑦)𝑑𝑦 = 0 𝑃(𝑥; 𝑦) = 3𝑥 2 𝑦 + 2 cos 𝑦 → 𝑃𝑦′ = 3𝑥 2 − 2 sin 𝑦 𝑄(𝑥; 𝑦) = 𝑥 3 − 2𝑥𝑠𝑖𝑛 𝑦 → 𝑄𝑥′

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