Bai tap mau 1: Tinh cac gioi han sau:5n3-3n2+6 a.lim_(n→∞)(5n^3-3n^2+6)/(4n^2-3n^3+7n) c.lim_(n→∞)(2n^2-n+3)/(3n^2+2n+1) b. lim_(n→∞)(6n^4+2n^2-1)/(1+5n^2-3n^4)1+5n2-3n d. lim(-2n^2+3n-1)/(n^2-1)lim_(√(n+n)^++1+n)...
Bai tap tu luyen: Tinh cac tich phan∫_0^(π/(2))(cosxdx)/(2sinx+1) ∫(cos^3x)/(sin^2x)dx ∫(cosx)/((1-sin^2x)sin^2x)dx ∫(sin4x)/(3+cos2x)dx tan' xdx∫(sin2xcosx)/(1+cosx)dx(DHKB-2005) dx (DHKB-2005) 相关知识点: 试题来源: 解析 a.b.c.g. 反...
Bai tap twong tr: Giài cac phuong trinh sau:1) 2x^2-6x-1=√(4x+5) .2) (x+3)√(48-8x-x^2)=x-24 .3 √(x^2-x-2)-2√(x-2)+2=√(x+1)4)√(x^2+10x+21)=3√(x+3)+2√(x+7)-6 .5)√(x^2+2x+4)+√(2-x)=√(8-x^3)+1 6) √x+√(x+1)-√(...
Bai tap 2.13. Giai cac phuong trinh sau:1) (x+1)⋅√(x+2)+(x+6)⋅√(x+7)=(x+3)⋅(x+4) ;2) (x+2)⋅√(x+3)+(x+7)⋅√(x+8)=(x+4)⋅(x+5) ;3) (x+4)⋅√(x+5)+(x+9)⋅√(x+10)=(x+6)⋅(x+7) ;4) (x+5)⋅√(x+6)+(x+10...
Bai tap 1: Tim cac gioi han sau a) lim_(x→0)_(x→0)(tanx-x)/(x-sinx)b) lim_(x→0)_(x→0)(x^3)/(x-sinx) 相关知识点: 试题来源: 解析 http://gallery.fbcontent.cn/latex?decode=false&latex=%24%24%5Cbegin%7Baligned%7D%20%26amp%3B%20%5Ctext%20%7B%20a)%20...
百度试题 结果1 题目Bai tap 92. Tinh cac gidi han sau.a)lim√x3+3x2√x2-2x b)lim①++∞ 相关知识点: 试题来源: 解析 a) Ta cób) Ta có 反馈 收藏
B. BAI TAP UNG DUNG1Viet cach doc cac so sau:109:...n n 10...110:...101:.105:...102:...… ...108:...■103:...104:...106:...107:... 相关知识点: 试题来源: 解析 109:Một trăm linh chín110Một trăm mười101:Một trăm linh...
Bai tap1. Tim gioi han cua cac ham so sau:a)lim_(x→∞)(2x+sin^2x-5cos2x)/(x^2+3) b)lim x2cos;c)x0 X lim_(x→∞)(√(x+1)-√x)/xcos(√(x+1)+√x) 相关知识点: 试题来源: 解析 a) Ta co: sin^2x-5cos2x=11sin^2x-5(2x-5)/(x^2+3)≤(2x+sin^2x-5...
Bai 2.44: Hay tim cac tap B(8), B(12), va BC (8, 12) 相关知识点: 试题来源: 解析 http://gallery.fbcontent.cn/latex?decode=false&latex=%24%24B(8)%3D%5C%7B0%20%3B%208%20%3B%2016%20%3B%2024%20%5Cldots%5C%7D%24%24%25&fontSize=30http://gallery.fbcontent.cn/late...
百度试题 结果1 题目Bai toan 9: Cho tam giac ABC. Tim tap hop cac diém M thoa man mot trong cac dieu kien sau: 相关知识点: 试题来源: 解析 a) là điểm tùy ý.b) là trọng tâm của tam giác . 反馈 收藏 ...