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

x -1 0 0-|||-0 x -1 0-|||-0 0 x -1 =a3x ^3+a 2x^2+a1x+a 0.-|||-a0 a1 a2 a3

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)_(

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

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-|

16.线性方程组 ) (x)_(1)+(x)_(3)=0 2(x)_(2)+(x)_(3)=0 2(x)_(1)+3(x)_(2)=0 .16.线性方程组用

已知集合A=(x∈R|3x+2＞0)，B=(x∈R|（x+1）（x-3）＞0)，则A∩B=（ ）A. （3，+∞）B. $（-1，-\frac{2}{3}）$C