2.用适当的符号填空:-|||-(1)a __ a,b,c ;;-|||-(2)0 __ x|{x)^2=0} ;-|||-(3)times __ xin R|{x)^2+1=0} ;-|||-(4) 0,1 __ N;-|||-(5)(0) __ x|{x)^2=x} ;-|||-(6)(2,1) __ x|{x)^2-3x+2=0} .

若_(1)+((k)^2+1)(x)_(2)+2(x)_(3)=0-|||-_(1)+(2k+1)(x)_(2)+2(x)_(3)=0-|||-(x)_(1)+

的解为 () .-|||-(A) x=2 =0, z=-2 ; (B) x=-2 =2 =0;-|||-(C) =0, =2, =-2; (D) x=1 =0,

若方程组 ) 7(x)_(1)+8(x)_(2)+9(x)_(3)=0 -(x)_(2)+2(x)_(3)=0 2(x)_(2)+t(x)_(3)=0 .=

已知lim _(xarrow +infty )(x)^a[ sqrt ({x)^2+1}+sqrt ({x)^2-1}-2x] =bneq 0 x]=b≠0，则

函数(x)=(({x)^2+1)}^2,则(x)=(({x)^2+1)}^2(x)=(({x)^2+1)}^2函数,则

函数(x)=(({x)^2+1)}^2,则(x)=(({x)^2+1)}^2(x)=(({x)^2+1)}^2函数,则

6、极限 lim _(xarrow +infty )(sqrt ({x)^2+x}-sqrt ({x)^2+1})= () 。(较难)-|||-A、0 B、 d

微分方程((y)^2+2)dx+y((x)^2+1)dy=0的通解为( )((y)^2+2)dx+y((x)^2+1)dy=0((y)^2+2

2.试用拉普拉斯定理计算行列式-|||-1 1 1 0 0-|||-1 2 3 0 0-|||-D= 0 1 1 1 1-|||-0 x1 x2 x3 x4-|

已知线性方程组 ) a(x)_(1)+(x)_(3)=1 (x)_(1)+a(x)_(2)+(x)_(3)=0 (x)_(1)+2(x)_(2)+a(x)_(