中,
的辖域为( ),
的辖域为( )。



谓词公式forall x)(P(x)arrow R(x,y))cap Q(x,y)可换名为forall x)(P(x)arrow R(x,y))cap Q(x,
[单选题]已知集合P={x|0 ≤x ≤5,x∈Z},Q={y|y=|x2-1|,x∈P},则P∩Q中元素的个数是( ).(A)3.(B)6.(C)8.(D)9.
lim _(x arrow 1) (x-y)/(x^2)-y^(2)=A. 0 ;B. $\frac{1}{2}$ ;C. $\infty$ ;D. 不存在.
5.根据函数极限或无穷大定义,填写下表:-|||-(x)arrow A (x)arrow infty (x)arrow +infty (x)arrow -i
4.证明:(1) lim _((x,y)arrow (0,0))dfrac (x+y)(x-y) 不存在;(2)-|||-lim _((x,y)arrow (0
【单选题】已知谓词公式(∀x)(∀y)(P(x, y)→Q(x, y)),将其化为子句集的结果正确的是A. S = {¬P(x,y)∨Q(x,y)}B. S =
求函数的极限 lim _(x arrow 2) (tan x y)/(y)=A. 0B. 2C. 不存在D. $\infty$
[单选题]已知y1(x)与y2(x)是方程y″+P(x)y′+Q(x)y=0的两个线性无关的特解,Y1(x)和Y2(x)分别是是方程y″+P(x)y′+Q(x)y=R1(x)和y″+P(x)y′+Q(x)y=R2(x)的特解。那么方程y″+P(x)y′+Q(x)y=R1(x)+R2(x)的通解应是:()A . c1y1+c2y2B . c1Y1(x)+c2Y2(x)C . c1y1+c2y2+Y1(x)D . c1y1+c2y2+Y1(x)+Y2(x)
【单选题】谓词公式 x(P(x)∨ yR(y))→Q(x)中量词 x的辖域是()。A. x(P(x)∨ yR(y))B. P(x)C. (P(x)∨ yR(y)
lim_(x arrow + infty)(x^2+y^2)sin (3)/(x^2)+y^(2)A. 0B. 3C. 1/3D. ∞